Question

1. based on the following balanced equation,

$2\text{fe}(s) + 3\text{cuso}_4(aq) \
ightarrow \text{fe}_2(\text{so}_4)_3(aq)+ 3\text{cu}(s)$
a. what is the charge on each iron in the product $\text{fe}_2(\text{so}_4)_3$
b. if 0.5206 g of iron is reacted, how many moles of copper should be produced?

Explanation:

Step1: Find Fe charge in $\text{Fe}_2(\text{SO}_4)_3$

Sulfate ion ($\text{SO}_4^{2-}$) has charge $-2$. Let charge of Fe be $x$.
$$2x + 3(-2) = 0$$
$$2x - 6 = 0 \implies x = +3$$

Step2: Calculate moles of Fe

Molar mass of Fe = $55.85\ \text{g/mol}$.
$$n_{\text{Fe}} = \frac{0.5206\ \text{g}}{55.85\ \text{g/mol}} \approx 0.009321\ \text{mol}$$

Step3: Relate Fe to Cu via mole ratio

From balanced equation, $2\ \text{mol Fe}
ightarrow 3\ \text{mol Cu}$.
$$n_{\text{Cu}} = 0.009321\ \text{mol Fe} \times \frac{3\ \text{mol Cu}}{2\ \text{mol Fe}}$$
$$n_{\text{Cu}} \approx 0.01398\ \text{mol}$$

Answer:

a. +3 (Iron(III) ion, $\text{Fe}^{3+}$)
b. $0.01398$ moles