Chem 101 problem, please help I just can’t figure it out

Question

What volume in mL of CO gas at STP are required to produce 10.4g of CH3COOH via a carbonylation reaction with excess CH3OH (methanol) CO(g) + CH3OH(l) = CH3COOH(l)

Answer 1

CO(g) + CH₃OH(l) = CH₃COOH(l) ... balanced equation

molar mass CH3COOH = 60.1 g / mol

From the balanced equation, we see that 1 mol CH₃COOH requires 1 mol CO.

So we will find the mols CH3COOH which will also be the mols CO needed. We will then convert mols CO to mls of CO

10.4 g CH₃COOH x 1 mol / 60.1 g = 0.173 moles

moles CO(g) needed = 0.173 mols CH₃COOH x 1 mol CO / mol CH₃COOH = 0.173 mols CO

At STP 1 mole of CO = 22.4 L (assuming ideal gas behavior), therefore...

0.173 mols CO x 22.4 L / mol = 3.88 L needed

3.88 L x 1000 mls / L = 3880 mls CO needed