Determine the yield of the reaction product if the reaction of 12 g of methanol and 10 g of acetic acid formed 10.5 g of ether.

Let’s execute the solution:
1. We write down the equation according to the problem statement:
m = 12 g. m = 10 g. m = 10.5 g;
CH3OH + CH3COOH = CH3COO – CH3 + H2O – esterification, methyl acetate was obtained;
2. Calculation by formulas:
M (CH3OH) = 32 g / mol;
M (CH3COOH) = 60 g / mol;
M (ether) = 74 g / mol;
Y (CH3OH) = m / M = 12/32 = 0.375 mol (substance in excess);
Y (CH3COOH) = m / M = 10/60 = 0.166 mol (deficient substance);
Y (ether) = 0.166 mol since the amount of these substances according to the equation is 1 mol.
3. Find the theoretical mass of the product, W:
m (ether) = Y * M = 0.166 * 74 = 12.3 g;
W = m (practical) / m (theoretical) * 100;
W (ether) = 10.5 / 12.3 * 100 = 85.36%
Answer: The yield of methyl acetate is 85.36%.