Gram to moles to moles to Grams pg 8

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Gram to moles to moles to Grams pg 8 (Use molar mass—mole ratio—molar mass) 23 g CO ?g Fe 1. For this reaction: Fe3O4 + 4 CO  3 Fe + 4 CO2 a. How many grams of Fe are produced from 23 grams of CO? 23 g CO 1 mol CO 28.0 g CO = 0.82 mol CO 0.82 mol CO 3 mol Fe 4 mol CO = 0.62 mol Fe 55.8 g Fe 1 mol Fe = 34.6 g Fe 0.62 mol Fe

= = Fe3O4 + 4 CO  3 Fe + 4 CO2 3 mol Fe 4 mol CO 167.4 g Fe 112 g CO 3 x 55.8 = 167.4 g Fe 4 x 28 = 112 g CO 167.4 g Fe 112 g CO = ? g Fe 23 g CO = 34.6 g Fe

0.95 g Fe3O4 1 mol Fe3O4 231.4 g Fe3O4 = 0.0041 mol Fe3O4 ? g CO2 For this reaction: Fe3O4 + 4 CO  3 Fe + 4 CO2 b. How many grams of CO2 are produced from 0.95 grams of Fe3O4? 0.95 g Fe3O4 1 mol Fe3O4 231.4 g Fe3O4 = 0.0041 mol Fe3O4 0.0041 mol Fe3O4 4 mol CO2 1 mol Fe3O4 = 0.0164 mol CO2 0.0164 mol CO2 44.0 g CO2 1 mol CO2 = 0.72 g CO2

= = 1 mol Fe3O4 4 mol CO2 231.4 g Fe3 O4 176 g CO2 For this reaction: Fe3O4 + 4 CO  3 Fe + 4 CO2 b. How many grams of CO2 are produced from 0.95 grams of Fe3O4? 1 mol Fe3O4 4 mol CO2 = 231.4 g Fe3 O4 176 g CO2 1 x [3 x 55.8 + 4 x 16] 4 x [1 x 12 + 2 x 16] 231.4 g Fe3O4 176 g CO2 = 0.95 g Fe3O4 ? g CO2 = 0.72 g CO2

7.85 g PbO 1 mol PbO 223.2 g PbO = 0.0352 mol PbO 0.0352 mol PbO 2. For this reaction: 6 PbO + O2  2 Pb3O4 a. How many grams of Pb3O4 are produced from 7.85 grams of PbO? 7.85 g PbO 1 mol PbO 223.2 g PbO = 0.0352 mol PbO 0.0352 mol PbO 2 mol Pb3O4 6 mol PbO = 0.0117 mol Pb3O4 0.0117 mol Pb3O4 685.6 g Pb3O4 1 mol Pb3O4 = 8.0 g Pb3O4

= = 6 mol PbO 2 mol Pb3O4 1339.2 g PbO 1371.2 g Pb3O4 6 x [207.2 + 16] 2. For this reaction: 6 PbO + O2  2 Pb3O4 a. How many grams of Pb3O4 are produced from 7.85 grams of PbO? 6 mol PbO 2 mol Pb3O4 = 1339.2 g PbO 1371.2 g Pb3O4 6 x [207.2 + 16] 2 x [3 x 207.2 + 4 x 16] 1339.2 g PbO 1371.2 g Pb3O4 = 7.85 g PbO ? g Pb3O4 = 8.0 g Pb3O4

1.75 g O2 1 mol O2 32.0 g O2 = 0.0547 mol O2 0.0547 mol O2 6 mol PbO 2. For this reaction: 6 PbO + O2  2 Pb3O4 b. How many grams of PbO must react with 1.75 grams of O2? 1.75 g O2 1 mol O2 32.0 g O2 = 0.0547 mol O2 0.0547 mol O2 6 mol PbO 1 mol O2 = 0.328 mol PbO 0.328 mol PbO 223.2 g PbO 1 mol PbO = 73.2 g PbO

= = 6 mol PbO 1 mol O2 1339.2 g PbO 32.0 g O2 6 x [207.2 + 16] 2. For this reaction: 6 PbO + O2  2 Pb3O4 b. How many grams of PbO must react with 1.75 grams of O2? 6 mol PbO 1 mol O2 = 1339.2 g PbO 32.0 g O2 6 x [207.2 + 16] [2 x 16] 1339.2 g PbO 32.0 g O2 = ? g PbO 1.75 g O2 = 73.2 g PbO

17 g Al 1 mol Al 27.0 g Al = 0.63 mol Al 0.63 mol Al 2 mol Al2O3 3. For this reaction: 4 Al + 3 O2  2 Al2O3 a. How many grams of Al2O3 will be formed from 17 grams of Al reacting? 17 g Al 1 mol Al 27.0 g Al = 0.63 mol Al 0.63 mol Al 2 mol Al2O3 4 mol Al = 0.31 mol Al2O3 0.31 mol Al2O3 102.0 g Al2O3 1 mol Al2O3 = 31.6 g Al2O3

= = 4 mol Al 2 mol Al2O3 108.0 Al 204.0g Al2O3 4 x [27.0] 3. For this reaction: 4 Al + 3 O2  2 Al2O3 a. How many grams of Al2O3 will be formed from 17 grams of Al reacting? 4 mol Al 2 mol Al2O3 = 108.0 Al 204.0g Al2O3 4 x [27.0] 2 x [2 x 27 +3 x 16] 108.0 g Al 204.0 g Al2O3 = 17 g Al ? g Al2O3 = 32.1 g Al2O3

12 g Al 1 mol Al 27.0 g Al = 0.44 mol Al 0.44 mol Al 3 mol O2 4 mol Al 3. For this reaction: 4 Al + 3 O2  2 Al2O3 b. How many grams of O2 are needed to react with 12 grams of Al? 12 g Al 1 mol Al 27.0 g Al = 0.44 mol Al 0.44 mol Al 3 mol O2 4 mol Al = 0.33 mol O2 0.33 mol O2 32.0 g O2 1 mol O2 = 10.6 g O2

= = 4 mol Al 3 mol O2 108.0 g Al 96.0 g O2 4 x [27.0] 3 x [2 x 16] 3. For this reaction: 4 Al + 3 O2  2 Al2O3 b. How many grams of O2 are needed to react with 12 grams of Al? 4 mol Al 3 mol O2 = 108.0 g Al 96.0 g O2 4 x [27.0] 3 x [2 x 16] 108.0 g Al 96.0 g O2 = 12 g Al ? g O2 = 10.7 g O2

1.25 g NH3 1 mol NH3 17.0 g NH3 = 0.0735 mol NH3 0.0735 mol NH3 4. For this reaction: 4 NH3 + 5 O2  4 NO + 6 H2O a. How many grams of O2 are needed to react with 1.25 grams of NH3? 1.25 g NH3 1 mol NH3 17.0 g NH3 = 0.0735 mol NH3 0.0735 mol NH3 5 mol O2 4 mol NH3 = 0.092 mol O2 0.092 mol O2 32.0 g O2 1 mol O2 = 2.9 g O2

= = 4 mol NH3 5 mol O2 68.0 g NH3 160.0 g O2 4 x [1 x 14 + 3 x 1] 4. For this reaction: 4 NH3 + 5 O2  4 NO + 6 H2O a. How many grams of O2 are needed to react with 1.25 grams of NH3? 4 mol NH3 5 mol O2 = 68.0 g NH3 160.0 g O2 4 x [1 x 14 + 3 x 1] 5 x [2 x 16] 68.0 g NH3 160.0 g O2 = 1.25 g NH3 ? g O2 = 2.9 g O2

7.65 g O2 1 mol O2 32.0 g O2 = 0.239 mol O2 0.239 mol O2 6 mol H2O 4. For this reaction: 4 NH3 + 5 O2  4 NO + 6 H2O b. How many grams of H2O are produced from 7.65 grams of O2? 7.65 g O2 1 mol O2 32.0 g O2 = 0.239 mol O2 0.239 mol O2 6 mol H2O 5 mol O2 = 0.287 mol H2O 0.287 mol H2O 18.0 g H2O 1 mol H2O = 5.2 g H2O

= = 6 mol H2O 5 mol O2 108.0 g H2O 160.0 g O2 6 x [2 x 1 + 1 x 16] 4. For this reaction: 4 NH3 + 5 O2  4 NO + 6 H2O b. How many grams of H2O are produced from 7.65 grams of O2? 6 mol H2O 5 mol O2 = 108.0 g H2O 160.0 g O2 6 x [2 x 1 + 1 x 16] 5 x [2 x 16] 108.0 g H2O 160.0 g O2 = ? g H2O 7.65 g O2 = 5.2 g H2O