Certainly! Let's solve this problem step-by-step.
### Step 1: Write the Balanced Reaction Equation
First, we balance the chemical equation for the reaction. The initial unbalanced reaction is:
[tex]\[ \text{FePO}_4 + \text{Na}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + \text{Na}_3\text{PO}_4 \][/tex]
Balancing it, we get:
[tex]\[ 2 \text{FePO}_4 + 3 \text{Na}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 2 \text{Na}_3\text{PO}_4 \][/tex]
### Step 2: Identify the Molar Masses
We need the molar masses of the compounds involved:
- Sodium sulfate ([tex]\(\text{Na}_2\text{SO}_4\)[/tex]): [tex]\(142.04 \ \text{g/mol}\)[/tex]
- Sodium phosphate ([tex]\(\text{Na}_3\text{PO}_4\)[/tex]): [tex]\(163.94 \ \text{g/mol}\)[/tex]
### Step 3: Convert Mass to Moles
Given mass of sodium sulfate is [tex]\(15 \ \text{g}\)[/tex]. Convert this mass to moles:
[tex]\[ \text{moles of Na}_2\text{SO}_4 = \frac{15 \ \text{g}}{142.04 \ \text{g/mol}} \approx 0.1056 \ \text{moles} \][/tex]
### Step 4: Use Stoichiometry to Find Moles of Sodium Phosphate
According to the balanced equation, the mole ratio of [tex]\(\text{Na}_2\text{SO}_4\)[/tex] to [tex]\(\text{Na}_3\text{PO}_4\)[/tex] is 3:2. Therefore, we can calculate the number of moles of [tex]\(\text{Na}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{moles of Na}_3\text{PO}_4 = \left(0.1056 \ \text{moles of Na}_2\text{SO}_4 \times \frac{2 \ \text{moles of Na}_3\text{PO}_4}{3 \ \text{moles of Na}_2\text{SO}_4}\right) \approx 0.0704 \ \text{moles} \][/tex]
### Step 5: Convert Moles of Sodium Phosphate to Mass
Now, convert the moles of sodium phosphate to grams:
[tex]\[ \text{mass of Na}_3\text{PO}_4 = 0.0704 \ \text{moles} \times 163.94 \ \text{g/mol} \approx 11.54 \ \text{g} \][/tex]
### Step 6: Adjust for Yield
The reaction yield is [tex]\(65.0 \%\)[/tex]. We adjust the theoretical yield to find the actual yield:
[tex]\[ \text{mass of Na}_3\text{PO}_4 \text{ (adjusted)} = 11.54 \ \text{g} \times \frac{65.0}{100} \approx 7.50 \ \text{g} \][/tex]
Therefore, the mass of sodium phosphate ([tex]\(\text{Na}_3\text{PO}_4\)[/tex]) produced is approximately [tex]\(7.50 \ \text{g}\)[/tex].
