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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
pstnonsonjoku

The percent yield of dimethyl oxalate is 74.7%.

The equation of the reaction is;

H2C2O4 + 2CH3OH --------> C2O4(CH3)2 + 2H2O

Number of moles of H2C2O4 = 91.0 g/90 g/mol = 1.01 moles

Number of moles of methanol = 79.0 g/32 g/mol = 2.47 moles

Since 1 mole of oxalic acid requires 2 moles of methanol

x moles of oxalic acid requires 2.47 moles of methanol

x =  1 mole ×  2.47 moles/2 moles

x = 1.24 moles of oxalic acid

We can see that there is not enough oxalic acid hence oxalic acid is the limiting reactant.

1 mole of oxalic acid yields 1 mole of dimethyl oxalate

1.01 moles of oxalic acid yields 1.01 moles of dimethyl oxalate

Therefore;

Theoretical yield of dimethyl oxalate = 1.01 moles × 118 g/mol = 119.18 g

The percentage yield is obtained from;

% yield = actual yield/theoretical yield × 100/1

Actual yield of dimethyl oxalate = 89.0 g

% yield = 89.0 g/ 119.18 g × 100/1

% yield = 74.7%

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