Unit 8 Review – Mixed Practice Please Show Your Work 1. What is the coefficient for calcium in the balanced equation? 2 Ca + 5 NaOH → 4 Na + 2 Ca(OH)2 2. The burning of 20.0 g of acetylene (C2H2) produces what mass of carbon dioxide? You must write the equation, balance it and solve the mass-mass problem. a. Balanced equation: 2 C2H2 + 5 O2 4 CO2 + 2 H2O b. Mass-mass solution: 20 g C2H2 ÷ 26.02 g/mol C2H2 × (4 mol CO2 / 2 mol C2H2) × 44.01 g/mol CO2 = 67.7 g CO2 3. For the reaction HCl + KOH → KCl + H2O, identify the limiting reactant given 35.0 g of potassium hydroxide and 35.0 g of hydrochloric acid? 35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O 35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O 35.0 g KOH is the limiting reactant 4. Given four substances labeled W, X, Y and Z in the following reaction, W + X → Y + Z, if substance W is in excess, what is the limiting reactant? X is the limiting reactant 5. The following balanced equation represents the burning of butane. 2C4H10 (g) + 13O2 (g)→ 8CO2(g) + 10H2O(g) How many moles of carbon dioxide are produced when 3 moles of butane are reacted? 3 mol C4H10 × (8 mol CO2 / 2 mol C4H10) = 12 mol CO2 6.a. A piece of magnesium burns in the presence of oxygen, forming magnesium oxide. Write the equation and balance it. 2 Mg + O2 2 MgO b. From the reaction in the previous question, how many moles of oxygen are needed to completely react with 12 moles of magnesium? 12 mol Mg × (1 mol O2 / 2 mol Mg) = 6 mol O2 c. If you begin with 25.0 g of magnesium, how many moles of magnesium oxide will be produced? 25 g Mg ÷ 24.3 g/mol Mg × (2 mol MgO / 2 mol Mg) = 1.03 mol MgO d. Convert the answer you got in 6c above into grams (mole B to gram B). Is that the actual yield or theoretical yield? 1.03 mol MgO × 40.3 g/mol MgO = 41.5 g MgO This is the theoretical yield 7. Hydrochloric acid reacts with potassium hydroxide. HCl + KOH → KCl + H2O If the elements are present in the correct ratios to completely react all of the participants, how many moles of potassium chloride will be formed from 8 moles of HCl? 8 mol HCl × (1 mol KCl / 1 mol HCl) = 8 mol KCl 8. How many moles of fluorine gas are in 50.0 g of F2 (g)? 50 g F2 ÷ 37.98 g/mol F2 = 1.32 mol F2 Unit 8 Review – Mixed Practice Please Show Your Work 9. In the reaction 4NH3 (aq) + 5O2 (g) → 4NO (g) + 6H2O (l), what is the mole ratio of ammonia to water? 4 : 6 10. Iron metal (Fe) can be obtained from iron ore (Fe2O3) by the following reaction, Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g) How many grams of iron ore are needed to obtain 92.8 g of iron metal? 92.8 g Fe ÷ 55.84 g/mol Fe × (1 mol Fe2O3 / 2 mol Fe) × 159.68 g/mol Fe2O3 = 133 g Fe2O3 11. If you burn 2.5 moles of methane (CH4) according to the following reaction, CH4 (g) +2 O2 (g) → CO2 (g) +2 H2O (g) a. How many grams of carbon dioxide are produced? 2.5 mol CH4 × (1 mol CO2 / 1 mol CH4) × 44.01 g/mol CO2 = 110 g CO2 b. What is the mole ratio of methane to carbon dioxide? 1 : 1 12. Aspirin (C9H8O4) can be made from salicylic acid (C7H6O3) and acetic anhydride (C4H6O3). Suppose you mix 13.2 g of salicylic acid with an excess of acetic anhydride and obtain 5.9 g of aspirin and some water. Calculate the percent yield of aspirin in this reaction. The balanced equation has been provided for you. 2 C7H6O3 (s) + C4H6O3 (l) → 2 C9H8O4 (s) + H2O (l) First calculate the theoretical yield. 13.2 g C7H6O3 ÷ 138.07 g/mol C7H6O3 × (2 mol C9H8O4 / 2 mol C7H6O3) × 180.09 g/mol C9H8O4 = 17.2 g C9H8O4 Now calculate percent yield. actual 5.9 g C H O % yield = × 100 = 9 8 4 × 100 = 34% theoretical 17.2 g C9H8O4 13. In an experiment, 10.0 g of magnesium reacted with excess hydrochloric acid forming magnesium chloride. Mg(s) + 2HCl(aq) → MgCl2 + H2(g) At the completion of the reaction, 29.5 g of magnesium chloride was produced. Calculate the theoretical yield and the percent yield. 10.0 g Mg ÷ 24.30 g/mol Mg × (1 mol MgCl2 / 1 mol Mg) × 95.2 g/mol MgCl2 = 39.2 g MgCl2 actual 29.5 g MgCl % yield = × 100 = 2 × 100 = 75.3% theoretical 39.2 g MgCl2 .