Stoikiometri 5j2s1a
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Stoichiometry Additional Examples
Stoichiometry Flow Chart mL soln A (M/1000) \
mL soln B / (1000/M)
Y moles B \ --------------/ 1/MM A \ X moles A / MM B g A ---------------> mol A -----------------> mol B ---------------> g B / \ (P/1000RT) / (1000RT/P) \ / \ mL A gas mL B gas where
M => molarity of solution MM => molar mass P => pressu re of gas R => gas constant T => temperature of gas
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) #g H2 = --------------------------------------------
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) #g H2 = -------------------------------------------(1 mol O2)
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) #g H2 = -------------------------------------------(1 mol O2)
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) (2.0 g H2) #g H2 = -------------------------------------------(1 mol O2) (1 mol H2) the molar mass of H2
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) (2.0 g H2) #g H2 = -------------------------------------------(1 mol O2) (1 mol H2) = 13 g H2
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) #g H2 = --------------------
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) #g H2 = --------------------------------------------(18.0 g H2O) molar mass of H2O
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) #g H2 = --------------------------------------------(18.0 g H2O)
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) #g H2 = --------------------------------------------(18.0 g H2O) (2 mol H2O)
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2) #g H2 = --------------------------------------------------------------(18.0 g H2O) (2 mol H2O) (1 mol H2) molar mass H2
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2) #g H2 = --------------------------------------------------------------(18.0 g H2O) (2 mol H2O) (1 mol H2) = 0.606 g H2
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ?
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ?
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ? ? => KCl KClO3 -----> O2 + KCl
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ? ? => KCl KClO3 -----> O2 + KCl 2 KClO3 -----> 3 O2 + 2 KCl
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass molar mass
= 39.2 g O2
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia N2 + H2 -----> NH3
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia N2 + H2 -----> NH3 N2 + 3 H2 -----> 2 NH3
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) #mol H2 = -----------------
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) #mol H2 = -----------------------------(1 kg)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) #mol H2 = -------------------------------------------(1 kg) (17.0 g NH3)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2) #mol H2 = ------------------------------------------------------------(1 kg) (17.0 g NH3) (2 mol NH3)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2) #mol H2 = ------------------------------------------------------------(1 kg) (17.0 g NH3) (2 mol NH3) = 2.6 X 102 mol H2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O2 -----> ZnO + SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide ZnS + O2 -----> ZnO + SO2 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2) #g O2 = ---------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2) #g O2 = ---------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
= 3.45 g O2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2) #gSO2 = --------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2) #gSO2 = --------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol SO2) = 4.60 g SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? #g O2 = 3.45 g O2 #gSO2 = 4.60 g SO2
EXAMPLE Zinc and sulfur react to form zinc sulfide, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is Zn + S -----> ZnS In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) Which is the limiting reactant? (b) How many grams of ZnS can be formed from this particular reaction mixture? (c) How many grams of which element will remain unreacted in this experiment?
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
= 17.9 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------(32.06 gS ) (1 mol S) (1 mol ZnS)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------(32.06 gS ) (1 mol S) (1 mol ZnS) = 19.8 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced.
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) How many grams of which element will remain unreacted in this experiment? if use all 12.0 g Zn
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S) #g S = -------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol S)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S) #g S = -------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol S) = 5.88 g S reacted
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn #g S = 5.88 g S reacted Therefore, (6.50 - 5.88)g = 0.61 g S remain unreacted.
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM
1(12.011) %C = -------------- X 100 = 10.061% C 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(1.00797) %H = ---------------- X 100 = 0.844359% H 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 3(35.453) %Cl = -------------- X 100 = 89.095% Cl 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = 119.377amu %C = 10.061% C %H = 0.844359% H %Cl = 89.095% Cl
Stoichiometry Flow Chart mL soln A (M/1000) \
mL soln B / (1000/M)
Y moles B \ --------------/ 1/MM A \ X moles A / MM B g A ---------------> mol A -----------------> mol B ---------------> g B / \ (P/1000RT) / (1000RT/P) \ / \ mL A gas mL B gas where
M => molarity of solution MM => molar mass P => pressu re of gas R => gas constant T => temperature of gas
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water?
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) #g H2 = --------------------------------------------
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) #g H2 = -------------------------------------------(1 mol O2)
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) #g H2 = -------------------------------------------(1 mol O2)
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) (2.0 g H2) #g H2 = -------------------------------------------(1 mol O2) (1 mol H2) the molar mass of H2
EXAMPLE What mass of H2, in grams, will react with 3.3 mol O2 to produce water? 2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2) (2.0 g H2) #g H2 = -------------------------------------------(1 mol O2) (1 mol H2) = 13 g H2
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O?
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) #g H2 = --------------------
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) #g H2 = --------------------------------------------(18.0 g H2O) molar mass of H2O
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) #g H2 = --------------------------------------------(18.0 g H2O)
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) #g H2 = --------------------------------------------(18.0 g H2O) (2 mol H2O)
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2) #g H2 = --------------------------------------------------------------(18.0 g H2O) (2 mol H2O) (1 mol H2) molar mass H2
EXAMPLE What mass of H2, in grams, must react with excess O2 to produce 5.40 g H2O? 2 H2 + O2 -----> 2 H2O (5.40 g H2O) (1 mol H2O) (2 mol H2) (2.02 g H2) #g H2 = --------------------------------------------------------------(18.0 g H2O) (2 mol H2O) (1 mol H2) = 0.606 g H2
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ?
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ?
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ? ? => KCl KClO3 -----> O2 + KCl
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO3 -----> O2 + ? ? => KCl KClO3 -----> O2 + KCl 2 KClO3 -----> 3 O2 + 2 KCl
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2)
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass molar mass
EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO3 -----> 3 O2 + 2 KCl (100. g KClO3) (1 mol KClO3) (3 moles O2) (32.0 g O2) #g O2 = ------------------------------------------------------------------------------(122.6 g KClO3) (2 moles KClO3) (1 mol O2) molar mass molar mass
= 39.2 g O2
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3?
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia N2 + H2 -----> NH3
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? nitrogen + hydrogen -----> ammonia N2 + H2 -----> NH3 N2 + 3 H2 -----> 2 NH3
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) #mol H2 = -----------------
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) #mol H2 = -----------------------------(1 kg)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) #mol H2 = -------------------------------------------(1 kg) (17.0 g NH3)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2) #mol H2 = ------------------------------------------------------------(1 kg) (17.0 g NH3) (2 mol NH3)
EXAMPLE Ammonia, NH3, is prepared by the Haber process by the reaction of nitrogen gas with hydrogen gas. How many moles of hydrogen are needed to produce 3.0 kg of NH3? N2 + 3 H2 -----> 2 NH3 (3.0 kg NH3) (1000 g) (1 mol NH3) (3 mol H2) #mol H2 = ------------------------------------------------------------(1 kg) (17.0 g NH3) (2 mol NH3) = 2.6 X 102 mol H2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide
ZnS + O2 -----> ZnO + SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide ZnS + O2 -----> ZnO + SO2 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2) #g O2 = ---------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS) (3 mol O2) (32.0 g O2) #g O2 = ---------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol O2)
= 3.45 g O2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2) #gSO2 = --------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol SO2)
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O2 -----> 2 ZnO + 2 SO2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO2) (64.06 g SO2) #gSO2 = --------------------------------------------------------------(97.44 g ZnS) (2 mol ZnS) (1 mol SO2) = 4.60 g SO2
EXAMPLE The first step in obtaining elemental zinc from its sulfide ore involves heating it in air to obtain the oxide of zinc and sulfur dioxide. Calculate the mass of O2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? #g O2 = 3.45 g O2 #gSO2 = 4.60 g SO2
EXAMPLE Zinc and sulfur react to form zinc sulfide, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is Zn + S -----> ZnS In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) Which is the limiting reactant? (b) How many grams of ZnS can be formed from this particular reaction mixture? (c) How many grams of which element will remain unreacted in this experiment?
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol ZnS)
= 17.9 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------(32.06 gS ) (1 mol S) (1 mol ZnS)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------(32.06 gS ) (1 mol S) (1 mol ZnS) = 19.8 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced.
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) How many grams of which element will remain unreacted in this experiment? if use all 12.0 g Zn
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S) #g S = -------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol S)
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol S) (32.06 g S) #g S = -------------------------------------------------------(65.38 g Zn) (1 mol Zn) (1 mol S) = 5.88 g S reacted
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
(c) if use all 12.0 g Zn #g S = 5.88 g S reacted Therefore, (6.50 - 5.88)g = 0.61 g S remain unreacted.
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(gaw) %C = ------------ X 100 MM
1(12.011) %C = -------------- X 100 = 10.061% C 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 1(1.00797) %H = ---------------- X 100 = 0.844359% H 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3*35.453)amu = 119.377amu 3(35.453) %Cl = -------------- X 100 = 89.095% Cl 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = 119.377amu %C = 10.061% C %H = 0.844359% H %Cl = 89.095% Cl
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