Question

Theoretical yield of Fe2(SO3)4 if 20 g of FePO4 reacts with excess Na2SO4

Answer 1

Answer: 27 g of Fe2(SO4)3

Step-by-step explanation:

I'm assuming that there is an error in the original question: Fe2(SO3)4 is not a likely product. Indeed, a reactant of FePO4 would suggest that we are working with Fe atoms with +3 charge, however, the product Fe2(SO3)4 would have Fe atoms with a 0 charge (since SO3 is normally neutral), which is not consistent.

The correct product should be Fe2(SO4)3.

Proceeding under this corrected information:

First we must determine the balanced chemical equation. To do this, start by noting the relevant charges of the reactants:

FePO4: PO4 = -3 charge, Fe = +3 charge

Na2SO4: SO4 = -2 charge, Na = + 1 charge.

Then, in order to keep the charges balanced, the reaction must be:

`2 \ \text{FePO}_4 + 3 \ \text{Na}_2\text{SO}_4 \rightarrow 2\ \text{Na}_3\text{PO}_4 + \text{Fe}_2(\text{SO}_4)_3`

Next, we assume that we have excess Na2SO4. Then,

20 g FePO4
x 1 mol FePO4 / (150.82 g FePO4)
x 1 mol Fe2(SO4)3 / (2 mol FePO4)
x 399.88 g Fe2(SO4)3 / (1 mol Fe2(SO4)3)
--------------------------
26.51 g Fe2(SO4)3

which we report as 27 g Fe2(SO4)3 after applying significant figures.