Certainly! Let's solve this problem step-by-step:
1. Write down the balanced chemical equation:
The reaction between iron(III) phosphate and sodium sulfate is given by:
[tex]\[
2 \text{FePO}_4 + 3 \text{Na}_2\text{SO}_4 \rightarrow \text{Fe}_2\left(\text{SO}_4\right)_3 + 2 \text{Na}_3\text{PO}_4
\][/tex]
However, it appears there might be a typo in this equation. Let's use the correctly balanced equation:
[tex]\[
\text{FePO}_4 + 3 \text{Na}_2\text{SO}_4 \rightarrow \text{Fe}_2\left(\text{SO}_4\right)_3 + 2 \text{Na}_3\text{PO}_4
\][/tex]
2. Determine the molar masses of relevant compounds:
- Molar mass of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]:
[tex]\[
(2 \times 23) + 32 + (4 \times 16) = 46 + 32 + 64 = 142 \text{ g/mol}
\][/tex]
- Molar mass of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]:
[tex]\[
(3 \times 23) + 31 + (4 \times 16) = 69 + 31 + 64 = 164 \text{ g/mol}
\][/tex]
3. Calculate the moles of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] used:
Given mass of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] is 15 g.
[tex]\[
\text{Moles of Na}_2\text{SO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \text{ g}}{142 \text{ g/mol}} \approx 0.1056 \text{ moles}
\][/tex]
4. Use stoichiometry to determine the moles of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex] produced:
From the balanced equation, 3 moles of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] produce 2 moles of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex].
Using the ratio:
[tex]\[
\text{Moles of Na}_3\text{PO}_4 = \left( \frac{2}{3} \right) \times \text{Moles of Na}_2\text{SO}_4 \approx \left( \frac{2}{3} \right) \times 0.1056 \approx 0.0704 \text{ moles}
\][/tex]
5. Calculate the theoretical mass of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]:
[tex]\[
\text{Mass of Na}_3\text{PO}_4 (\text{theoretical}) = \text{moles} \times \text{molar mass} = 0.0704 \text{ moles} \times 164 \text{ g/mol} \approx 11.55 \text{ g}
\][/tex]
6. Calculate the actual mass of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex] produced based on the yield:
Given a yield of 65.0%:
[tex]\[
\text{Actual mass of Na}_3\text{PO}_4 = \left( \frac{65.0}{100} \right) \times 11.55 \text{ g} \approx 7.51 \text{ g}
\][/tex]
Summary:
- Moles of Na[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: [tex]\(0.1056\)[/tex]
- Moles of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]: [tex]\(0.0704\)[/tex]
- Theoretical mass of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]: [tex]\(11.55\)[/tex] g
- Actual mass of Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex] produced: [tex]\(7.51\)[/tex] g
So, [tex]\(7.51\)[/tex] grams of sodium phosphate (Na[tex]\(_3\)[/tex]PO[tex]\(_4\)[/tex]) will be produced under the given conditions.
