In the industrial synthesis of acetic acid, methanol is made to react with carbon monoxide. The balanced equation for this reaction is as follows: 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

First, calculate the molar mass of acetic acid (CH3COOH) which is approximately 60.05 g/mol. Given that 49.4 g of acetic acid is produced, we find the moles of acetic acid by dividing the mass by its molar mass: 49.4 g / 60.05 g/mol ≈ 0.822 moles. Since the reaction is 1:1 for CO to acetic acid, the moles of CO required are also 0.822. The molar mass of CO is approximately 28.01 g/mol, so the mass of CO required is 0.822 moles * 28.01 g/mol ≈ 23.0 g.

In the Industrial Synthesis of Acetic Acid, Methanol Is Made