If 91.0 g of oxalic acid and 79.0 g of methanol react to form 89.0 g of dimethyl oxalate, what is the percent yield?

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H2C2O4 + 2 CH3OH → (CH3)2(COO)2 + 2 H2O 

(91.0 g H2C2O4) / (90.0349 g H2C2O4/mol) = 1.0107 mol H2C2O4 
(79.0 g CH3OH) / (32.0419 g CH3OH/mol) = 2.4655 mol CH3OH 

(1.0107 moles of H2C2O4 would react completely with 1.0107 x (2/1) = 2.0214 moles of CH3OH, but there is more CH3OH present than that, so CH3OH is in excess and H2C2O4 is the limiting reactant. 

(1.0107 mol H2C2O4) x (1 mol (CH3)2(COO)2 / 1 mol H2C2O4) x (118.0880 g (CH3)2(COO)2/mol) = 
119.35 g (CH3)2(COO)2 in theory 

(89.0 g) / (119.35 g) = 0.746 = 74.6% yield

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