Problem #6 (20 pt)
nAluminum (III) sulfate reacts with sodium nitrate to give the double replacement products sodium sulfate and aluminum (III) nitrate.  If I have 11.00 g aluminum (III) sulfate , how many grams of sodium nitrate would I need for a complete reaction to occur? If, after complete reaction, I recover 1.50 g of aluminum (III) nitrate, what was the % yield of the reaction?
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nAl2(SO4)3  + 6  NaNO3 → 3 Na2SO4 + 2 Al(NO3)3
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n11.00 g Al2(SO4)3 * 1 mol Al2(SO4)3 * 6 mol NaNO3 * 85.0 g NaNO3 = 16.4 g NaNO3
n                       342.2 g Al2(SO4)3  1 mol Al2(SO4)3    1 mol NaNO3 
n11.00 g Al2(SO4)3 * 1 mol Al2(SO4)3 * 2 mol Al(NO3)3 * 213 g Al(NO3)3 = 13.6 g Al(NO3)3
n                       342.2 g Al2(SO4)3  1 mol Al2(SO4)3    1 mol  AL(NO3)3 
n1.50 g actual * 100 = 10.9% yield
n13.6 g theoretical