Question

The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon

monoxide
CHCOOH
CH3OH + CO
Relative atomie mass: H = 1:0 = 16; C = 12
Calculate the maximum mass of ethanoic acid that can be produced from 160g methanol,
assuming the carbon monoxide is in excess.

Answer 1

`\boxed{\text{300 g}}`

Step-by-step explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32 60

CH₃OH + CO ⟶ CH₃COOH

m/g: 160

(a) Moles of CH₃OH

`\text{Moles of CH$_(3)$OH} = \text{160 g CH$_(3)$OH }* \frac{\text{1 mol CH$_(3)$OH }}{\text{32 g CH$_(3)$OH}}= \text{5.00 mol CH$_(3)$OH}`

(b) Moles of CH₃COOH

`\text{Moles of CH$_(3)$COOH} = \text{5.00 mol CH$_(3)$OH } * \frac{\text{1 mol CH$_(3)$COOH}}{\text{1 mol CH$_(3)$OH }} = \text{5.00 mol CH$_(3)$COOH}`

(c) Mass of CH₃COOH

`\text{Mass of CH$_(3)$COOH} =\text{5.00 mol CH$_(3)$COOH} * \frac{\text{60 g CH$_(3)$COOH}}{\text{1 mol CH$_(3)$COOH}} = \textbf{300 g CH$_(3)$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}`