In the industrial synthesis of acetic acid, methanol is reacted with carbon monoxide. The balanced equation for this reaction is: CH₃OH(g) + CO(g) → CH₃COOH(l)
How many grams of CO(g) are required to produce 49.4 g of acetic acid?

Question

Quizplus · Accepted Answer

The molar mass of acetic acid (CH₃COOH) is 60.05 g/mol. Therefore, 49.4 g of acetic acid is 49.4 g / 60.05 g/mol = 0.822 mol. The reaction shows a 1:1 molar ratio between CO and CH₃COOH, so 0.822 mol of CO is needed. The molar mass of CO is 28.01 g/mol, so 0.822 mol × 28.01 g/mol = 23.0 g of CO is required.

In the Industrial Synthesis of Acetic Acid, Methanol Is Reacted