Aim # 12: How do we solve mass-mass problems? HW # 12 Study pp. 258-264 pp. 277-279 Study class notes Ans. ques. p. 282 # 11 p. 283 # 17,18,21,23,42

I Solving mass-mass problems Problem: Based upon the equation 3H2(g) + N2(g) → 2NH3(g) How many grams of ammonia are produced by the reaction of 56 g of nitrogen with excess hydrogen? Solution: For the reaction A+B→C+D mass x moles x moles x mass A ÷ A ÷ C ÷ C

Step 1. Change the given mass to moles using an appropriate conversion factor. e.g. 56g N2 x 1 mol N2 = 2.0 mol N2 28g N2 Step 2. Find the number of moles of the substance sought using coefficients from the balanced equation. e.g. 2.0 mol N2 x 2 mol NH3 = 4.0 mol NH3 1 mol N2 Step 3. Change the moles of the substance sought to mass using an appropriate conversion factor. e.g. 4.0 mol NH3 x 17 g NH3 = 68 g NH3 1 mol NH3

II Problem: How many grams of hydrogen gas are produced when enough nitric acid is added to 48.6g of magnesium to react it completely? Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g) Ans. 48.6 g Mg x 1 mol Mg = 2.00 mol Mg 24.3 g Mg 2.00 mol Mg x 1 mol H2 = 2.00 mol H2 1 mol Mg 2.00 mol H2 x 2.00 g H2 = 4.00 g H2 1 mol H2

III Percent Yield Yield is the mass of product obtained. % yield = actual yield x 100 theoretical yield In the previous problem, if we obtained 3.70 g H2, what would our % yield be? % yield = 3.70 g x 100 = 92.5% 4.00 g

Problem: For the reaction 2 C2H6 + 7O2 → 4CO2 + 6H2O How many grams of ethane (C2H6) need to be burned in order to produce 162 grams of water? Ans. 162 g H2O x 1 mol H2O = 9.00 mol H2O 18.0 g H2O 9.00 mol H2O x 2 mol C2H6 = 3.00 mol C2H6 6 mol H2O 3.00 mol C2H6 x 30.0 g C2H6 = 90.0 g C2H6 1 mol C2H6