Question

in order to have 0.349 mol of feso₄, what would this sample weigh?
○ 0.0325 g
○ 109 g
○ 72.5 g
○ 53.0 g

Explanation:

Step1: Calculate molar mass of $\ce{FeSO4}$

Molar mass of $\ce{Fe}$: $55.85\ \text{g/mol}$, $\ce{S}$: $32.07\ \text{g/mol}$, $\ce{O}$: $16.00\ \text{g/mol}$.
Molar mass of $\ce{FeSO4}$ = $55.85 + 32.07 + (4\times16.00)$ = $55.85 + 32.07 + 64.00$ = $151.92\ \text{g/mol}$ (approx $152\ \text{g/mol}$).

Step2: Use $m = n \times M$ (mass = moles × molar mass)

Given $n = 0.349\ \text{mol}$, $M = 151.92\ \text{g/mol}$.
$m = 0.349\ \text{mol} \times 151.92\ \text{g/mol} \approx 0.349 \times 152 \approx 53.0\ \text{g}$.

Answer:

53.0 g (the option with 53.0 g)