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The chemical equation shows iron(III) phosphate reacting with sodium sulfate.

[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]

What is the theoretical yield of [tex]\(\text{Fe}_2(\text{SO}_4)_3\)[/tex] if 20.00 g of [tex]\(\text{FePO}_4\)[/tex] reacts with an excess of [tex]\(\text{Na}_2\text{SO}_4\)[/tex]?

A. 26.52 g
B. 53.04 g
C. 150.8 g
D. 399.9 g


Sagot :

To determine the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex] from the given reaction, we will follow these steps:

### Step 1: Understand the Balanced Equation
The balanced chemical reaction is as follows:
[tex]\[ 2 \ FePO_4 + 3 \ Na_2SO_4 \rightarrow Fe_2(SO_4)_3 + 2 \ Na_3PO_4 \][/tex]

### Step 2: Identify Molar Masses
From the information provided:
- Molar mass of [tex]\( FePO_4 \)[/tex] = 150.82 g/mol
- Molar mass of [tex]\( Fe_2(SO_4)_3 \)[/tex] = 399.88 g/mol

### Step 3: Calculate Moles of [tex]\( FePO_4 \)[/tex]
Given mass of [tex]\( FePO_4 \)[/tex] = 20.00 g

Number of moles of [tex]\( FePO_4 \)[/tex] is calculated using the formula:
[tex]\[ \text{moles of } FePO_4 = \frac{\text{mass of } FePO_4}{\text{molar mass of } FePO_4} \][/tex]
[tex]\[ \text{moles of } FePO_4 = \frac{20.00 \ \text{g}}{150.82 \ \text{g/mol}} \approx 0.1326 \ \text{mol} \][/tex]

### Step 4: Use Stoichiometry to Find Moles of [tex]\( Fe_2(SO_4)_3 \)[/tex]
From the balanced equation:
[tex]\[ 2 \ \text{moles of } FePO_4 \rightarrow 1 \ \text{mole of } Fe_2(SO_4)_3 \][/tex]

Therefore, the moles of [tex]\( Fe_2(SO_4)_3 \)[/tex] produced:
[tex]\[ \text{moles of } Fe_2(SO_4)_3 = \frac{0.1326 \ \text{mol} \ FePO_4}{2} = 0.0663 \ \text{mol} \][/tex]

### Step 5: Calculate the Theoretical Yield (Mass) of [tex]\( Fe_2(SO_4)_3 \)[/tex]
The mass of [tex]\( Fe_2(SO_4)_3 \)[/tex] is:
[tex]\[ \text{mass of } Fe_2(SO_4)_3 = \text{moles of } Fe_2(SO_4)_3 \times \text{molar mass of } Fe_2(SO_4)_3 \][/tex]
[tex]\[ \text{mass of } Fe_2(SO_4)_3 = 0.0663 \ \text{mol} \times 399.88 \ \text{g/mol} \approx 26.52 \ \text{g} \][/tex]

### Final Answer
The theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex] when 20.00 g of [tex]\( FePO_4 \)[/tex] reacts with an excess of [tex]\( Na_2SO_4 \)[/tex] is approximately 26.52 g.