<p> Stoichiometry Review Show all work!</p><p>1. 4 Na + O2 ----> 2 Na2O a. How many moles of sodium would be needed to react with 3.82 moles of oxygen?</p><p>3.82 mol O2 x 4 mol Na = 15.3 mol Na 1 1 mol O2 b. How many moles of Na2O can be produced from 13.5 moles Na?</p><p>13.5 mol Na x 2 mol Na2O = 6.75 mol Na2O 1 4 mol Na c. How many moles of O2 are needed to produce 34.7 g of Na2O?</p><p>34.7 g Na2O x 1 mol Na2O x 1 mol O2 = 0.280 mol O2 1 61.98 g Na2O 2 mol Na2O 2. C2H4 + 3 O2 ---> 2 CO2 + 2 H2O a. When 0.624 moles of O2 are reacted, how many moles of carbon dioxide are produced?</p><p>0.624 mol O2 x 2 mol CO2 = 0.416 mol CO2 1 3 mol O2 b. How many grams of C2H4 are needed to produce 3.7 moles of water?</p><p>3.7 mol H2O x 1 mol C2H4 x 28.05 g C2H4 = 51.9 g C2H4 1 2 mol H2O 1 mol C2H4 c. How many grams of O2 are needed to react with 2.56 g of C2H4?</p><p>2.56 g C2H4 x 1 mol C2H4 x 3 mol O2 x 32.00 g O2 = 8.76 g O2 1 28.05 g C2H4 1 mol C2H4 1 mol O2 3. N2 + 3 F2 ---> 2 NF3 a. When 62.0 g of fluorine are reacted, how many moles of NF3 will be formed?</p><p>62.0 g F2 x 1 mol F2 x 2 mol NF3 = 1.09 mol NF3 1 38.00 g F2 3 mol F2 b. How many molecules of N2 are needed to produce 2.85 g of NF3? 23 22 2.85 g NF3 x 1 mol NF3 x 1 mol N2 x 6.02 x10 molecules N2 = 1.21 x 10 1 71.01 g NF3 2 mol NF3 1 mol N2 molecules N2 c. 3.54 g of nitrogen will react with how many grams of fluorine?</p><p>3.54 g N2 x 1 mol N2 x 3 mol F2 x 38.00 g F2 = 14.4 g F2 1 28.02 g N2 1 mol N2 1 mol F2 4. 4 NH3 + 7 O2 ---> 4 NO2 + 6 H2O a. What mass of NO2 can be produced from 3.56 x 1022 molecules of oxygen? 22 3.56 x 10 molecules O2 x 1 mol O2 x 4 mol NO2 x 46.01 g NO2 = 23 1 6.02 x 10 molecules O2 7 mol O2 1 mol NO2 1.55 g NO2 b. 13.8 g of NH3 would be able to produce how many moles of H2O?</p><p>13.8 g NH3 x 1 mol NH3 x 6 mol H2O = 1.22 mol H2O 1 17.03 g NH3 4 mol NH3 c. How many grams of O2 are needed to produce 15.5 g of H2O?</p><p>15.5 g H2O x 1 mol H2O x 7 mol O2 x 32.00 g O2 = 32.1 g O2 1 18.02 g H2O 6 mol H2O 1 mol O2 5. 2 Al + 6 HCl ----> 3 H2 + 2 AlCl3 </p><p> a. If you start with 5 L of 1.5 M HCl and you make 1.98 g of H2. What is your % yield? 1.5 M HCl = x / 5 L x = 7.5 mol HCl</p><p>7.5 mol HCl x 3 mol H2 x 2.02 g H2 = 7.575 g H2 1 6 mol HCl 1 mol H2 </p><p>(1.98 g H2 / 7.575 g H2) x 100% = 26.1%</p><p> b. Suppose your % yield was 88.3% instead, how much H2 would you make?</p><p>(x/ 7.575 g H2) x 100% = 88.3% x = 6.69 g H2 6. 2 Fe2S3 + 3 C ----> 4 Fe + 3 CS2 a. How many grams of iron can be made from 119 g of Fe2S3 and 12.7 g C?</p><p>119 g Fe2S3 x 1 mol Fe2S3 x 4 mol Fe x 55.85 g Fe = 63.9 g Fe 1 207.88 g Fe2S3 2 mol Fe2S3 1 mol Fe</p><p>12.7 g C x 1 mol C x 4 mol Fe x 55.85 g Fe = 78.7 g Fe 1 12.01 g C 3 mol C 1 mol Fe 63.9 g Fe b. What is the limiting reagent? </p><p>Fe2S3 c. After the above reaction, you produce 35.6 g of Fe. What is your % yield?</p><p>(35.6 g Fe / 63.9 g Fe) x 100% = 55.7% 7. 6 CO2 + 6 H2O ----> C6H12O6 + 6 O2</p><p> a. How many grams of sugar can be made from 50.0 g of CO2 and 50.0 g of water?</p><p>50.0 g CO2 x 1 mol CO2 x 1 mol C6H12O6 x 180.16 g C6H12O6 = 34.1 g C6H12O6 1 44.01 g CO2 6 mol CO2 1 mol C6H12O6</p><p>50.0 g H2O x 1 mol H2O x 1 mol C6H12O6 x 180.16 g C6H12O6 = 83.3 g C6H12O6 1 18.02 g H2O 6 mol H2O 1 mol C6H12O6 34.1 g C6H12O6</p><p> b. What is the limiting reagent?</p><p>CO2</p><p> c. What is the excess reagent?</p><p>H2O</p><p> d. How many grams of the excess reagent are leftover?</p><p>50.0 g CO2 x 1 mol CO2 x 6 mol H2O x 18.02 g H2O = 20.47261986 g H2O 1 44.01 g CO2 6 mol CO2 1 mol H2O</p><p>50.0 g H2O - 20.47261986 g H2O = 29.5 g H2O</p>