CHAPTER 11 CHEMISTRY

STOICHIOMETRY Review stoichiometry CH4(g) + 2O2(g) --> 2H2O(g) + CO2(g) If you had 5.0 moles of oxygen how many moles would you get of carbon dioxide? If you had 85 g of methane how many moles of water would you form? For every 5.0 grams of water how many grams of carbon dioxide would you have? If you have 45 g of oxygen how many molecules of carbon dioxide would you form?

LIMITING REACTANTS When one reactant has been completely used to form the product, and you still have some of the other reactant left over. 2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g) If 90.0 g of sodium is dropped into 80.0 g water, how many grams of H2 would be produced? What is the limiting reactant? How much of the other reactant was left over? 90.0gNa /2.0 g H2 / 46.0 g Na 3.90 g H2 80.0 gH2O / 2.0 g H2 / 36.0 g H2O 4.44 g H2 Sodium has to be the limiting reactant 90.0gNa / 36.0 g H2O = / 46.0g Na 70.4 so subtract 80.0 - 70.4 = 9.6 g H2O left over

PERCENT YIELD If you did an experiment where you start with 2.45 g Fe. Fe(s) + CuSO4(aq) --> FeSO4(aq) + Cu(s) How much copper did you form if you got a 73.4% yield? Actual yield X 100 = % yield Theoretical yield 2.45 g Fe / 63.5 g Cu = / 55.8 g Fe 2.79 g Cu X X 100 = 73.4% X = 73.4/100 X 2.79 X = 2.05 g Cu

MASS TO ENERGY Compare the heat of reaction for the displacement of 0.0663 g of sodium chloride from sodium bromide by chlorine Cl2(g) + 2 NaBr(aq) --> 2NaCl(aq) + Br2(l) ΔH = - 100.18 kJ Part of the product so you can use it in as a coefficient 0.0663g NaCl / 1 mol NaCl / -100.18 kJ = / 58.5 g NaCl / 2 mol -0.0568 kJ