Stoichiometry Additional Examples. HST Mr.Watson.

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Presentation transcript:

Stoichiometry Additional Examples

HST Mr.Watson

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water?

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O (3.3 mol O 2 ) #g H 2 =

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O (3.3 mol O 2 ) #g H 2 = (1 mol O 2 )

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O (3.3 mol O 2 ) (2 mol H 2 ) #g H 2 = (1 mol O 2 ) stoichiometric factor

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O (3.3 mol O 2 ) (2 mol H 2 ) (2.0 g H 2 ) #g H 2 = (1 mol O 2 ) (1 mol H 2 ) the molar mass of H 2

HST Mr.Watson EXAMPLE What mass of H 2, in grams, will react with 3.3 mol O 2 to produce water? 2 H 2 + O > 2 H 2 O (3.3 mol O 2 ) (2 mol H 2 ) (2.0 g H 2 ) #g H 2 = (1 mol O 2 ) (1 mol H 2 ) = 13 g H 2

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O?

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) #g H 2 =

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) (1 mol H 2 O) #g H 2 = (18.0 g H 2 O) molar mass of H 2 O

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) (1 mol H 2 O) (2 mol H 2 ) #g H 2 = (18.0 g H 2 O)

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) (1 mol H 2 O) (2 mol H 2 ) #g H 2 = (18.0 g H 2 O) (2 mol H 2 O) stoichiometric factor

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) (1 mol H 2 O) (2 mol H 2 ) (2.02 g H 2 ) #g H 2 = (18.0 g H 2 O) (2 mol H 2 O) (1 mol H 2 ) molar mass H 2

HST Mr.Watson EXAMPLE What mass of H 2, in grams, must react with excess O 2 to produce 5.40 g H 2 O? 2 H 2 + O > 2 H 2 O (5.40 g H 2 O) (1 mol H 2 O) (2 mol H 2 ) (2.02 g H 2 ) #g H 2 = (18.0 g H 2 O) (2 mol H 2 O) (1 mol H 2 ) = g H 2

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate.

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ?

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO > O 2 + ?

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO > O 2 + ? ? => KCl KClO > O 2 + KCl

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. potassium chlorate -----> oxygen + ? KClO > O 2 + ? ? => KCl KClO > O 2 + KCl 2 KClO > 3 O KCl

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO > 3 O KCl (100. g KClO 3 ) (1 mol KClO 3 ) (3 moles O 2 ) (32.0 g O 2 ) #g O 2 = (122.6 g KClO 3 ) (2 moles KClO 3 ) (1 mol O 2 )

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO > 3 O KCl (100. g KClO 3 ) (1 mol KClO 3 ) (3 moles O 2 ) (32.0 g O 2 ) #g O 2 = (122.6 g KClO 3 ) (2 moles KClO 3 ) (1 mol O 2 ) molar mass

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO > 3 O KCl (100. g KClO 3 ) (1 mol KClO 3 ) (3 moles O 2 ) (32.0 g O 2 ) #g O 2 = (122.6 g KClO 3 ) (2 moles KClO 3 ) (1 mol O 2 ) molar mass stoichiometric factor

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO > 3 O KCl (100. g KClO 3 ) (1 mol KClO 3 ) (3 moles O 2 ) (32.0 g O 2 ) #g O 2 = (122.6 g KClO 3 ) (2 moles KClO 3 ) (1 mol O 2 ) molar mass molar mass stoichiometric factor

HST Mr.Watson EXAMPLE Calculate the weight of oxygen produced during the thermal decomposition of 100. g of potassium chlorate. 2 KClO > 3 O KCl (100. g KClO 3 ) (1 mol KClO 3 ) (3 moles O 2 ) (32.0 g O 2 ) #g O 2 = (122.6 g KClO 3 ) (2 moles KClO 3 ) (1 mol O 2 ) molar mass molar mass = 39.2 g O 2 stoichiometric factor

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ?

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? nitrogen + hydrogen -----> ammonia

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? nitrogen + hydrogen -----> ammonia N 2 + H > NH 3

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? nitrogen + hydrogen -----> ammonia N 2 + H > NH 3 N H > 2 NH 3

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? N H > 2 NH 3 (3.0 kg NH 3 ) #mol H 2 =

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? N H > 2 NH 3 (3.0 kg NH 3 ) (1000 g) #mol H 2 = (1 kg)

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? N H > 2 NH 3 (3.0 kg NH 3 ) (1000 g) (1 mol NH 3 ) #mol H 2 = (1 kg) (17.0 g NH 3 )

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? N H > 2 NH 3 (3.0 kg NH 3 ) (1000 g) (1 mol NH 3 ) (3 mol H 2 ) #mol H 2 = (1 kg) (17.0 g NH 3 ) (2 mol NH 3 ) stoichiometric factor

HST Mr.Watson EXAMPLE Ammonia, NH 3, 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 NH 3 ? N H > 2 NH 3 (3.0 kg NH 3 ) (1000 g) (1 mol NH 3 ) (3 mol H 2 ) #mol H 2 = (1 kg) (17.0 g NH 3 ) (2 mol NH 3 ) = 2.6 X 10 2 mol H 2

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced?

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide ZnS + O > ZnO + SO 2

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? zinc sulfide + oxygen -----> zinc oxide + sulfur dioxide ZnS + O > ZnO + SO 2 2 ZnS + 3 O > 2 ZnO + 2 SO 2

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O > 2 ZnO + 2 SO 2 (7.00 g ZnS) (1 mol ZnS) (3 mol O 2 ) (32.0 g O 2 ) #g O 2 = (97.44 g ZnS) (2 mol ZnS) (1 mol O 2 ) stoichiometric factor

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O > 2 ZnO + 2 SO 2 (7.00 g ZnS) (1 mol ZnS) (3 mol O 2 ) (32.0 g O 2 ) #g O 2 = (97.44 g ZnS) (2 mol ZnS) (1 mol O 2 ) = 3.45 g O 2

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O > 2 ZnO + 2 SO 2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO 2 ) (64.06 g SO 2 ) #gSO 2 = (97.44 g ZnS) (2 mol ZnS) (1 mol SO 2 )

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? 2 ZnS + 3 O > 2 ZnO + 2 SO 2 (7.00 g ZnS) (1 mol ZnS)(2 mol SO 2 ) (64.06 g SO 2 ) #gSO 2 = (97.44 g ZnS) (2 mol ZnS) (1 mol SO 2 ) = 4.60 g SO 2

HST Mr.Watson 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 O 2 required to react with 7.00 g of zinc sulfide. How much sulfur dioxide is produced? #g O 2 = 3.45 g O 2 #gSO 2 = 4.60 g SO 2

HST Mr.Watson 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.

HST Mr.Watson 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?

HST Mr.Watson 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)

HST Mr.Watson 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

HST Mr.Watson 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)

HST Mr.Watson 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

HST Mr.Watson 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.

HST Mr.Watson 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

HST Mr.Watson 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)

HST Mr.Watson 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

HST Mr.Watson 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, ( )g = 0.61 g S remain unreacted.

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic?

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu = amu

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu = amu 1(gaw) %C = X 100 MM

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu = amu 1(gaw) %C = X 100 MM 1(12.011) %C = X 100 = % C

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu = amu 1( ) %H = X 100 = % H

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = ( *35.453)amu = amu 3(35.453) %Cl = X 100 = % Cl

HST Mr.Watson EXAMPLE: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? MM = 1(gaw) C + 1(gaw) H + 3(gaw) Cl = amu %C = % C %H = % H %Cl = % Cl