Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s)

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Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) What mass of silver is produced when 0.500 g of copper shavings are added to a silver nitrate solution containing 1.95 g of silver nitrate? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) 0.500 g Cu x 1 mol Cu x 2 mol Ag x 107.90 g Ag 63.55 g Cu 1 mol Cu 1 mol Ag = 1.70 g Ag 1.95 g AgNO3 x 1 mol AgNO3 x 2 mol Ag x 107.90 g Ag 169.91 g AgNO3 2 mol AgNO3 1 mol Ag = 1.24 g Ag

Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) How many grams of the excess reactant are left over when this reaction is complete? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) 1.24 g Ag x 1 mol Ag x 1 mol Cu x 63.55 g Cu = 0.365 g Cu 107.90 g Ag 2 mol Ag 1mol Cu 0.500g Cu - 0.365g Cu = 0.135 g Cu

If a student really performed this reaction in the lab and only collected 1.17g of silver, what would be her percent yield? Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s) % Yield = Actual Yield x 100 Theoretical Yield % Yield = 1.17 g Ag x 100 = 94.4% 1.24 g Ag

% Yield = Actual Yield x 100 Theoretical Yield 0.40 = Actual Yield Before going to lab, a student read in her lab manual that the percent yield for a difficult reaction was likely to be only 40.% of the theoretical yield. The student's prelab stoichiometric calculations predict that the theoretical yield should be 12.5 g. What's the student's likely actual yield? % Yield = Actual Yield x 100 Theoretical Yield 0.40 = Actual Yield 12.5 g Actual Yield = (0.40) (12.5g) = 5.0 g