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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 [tex]\(20.00 \, \text{g}\)[/tex] of [tex]\(\text{FePO}_4\)[/tex] reacts with an excess of [tex]\(\text{Na}_2\text{SO}_4\)[/tex]?

A. [tex]\(26.52 \, \text{g}\)[/tex]

B. [tex]\(53.04 \, \text{g}\)[/tex]

C. [tex]\(150.8 \, \text{g}\)[/tex]

D. [tex]\(399.9 \, \text{g}\)[/tex]

Sagot :

GinnyAnswer
To determine the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex] when 20.00 grams of [tex]\( FePO_4 \)[/tex] reacts with an excess of [tex]\( Na_2SO_4 \)[/tex], we should follow these steps:

1. Calculate the molar mass of [tex]\( FePO_4 \)[/tex]:
The molar mass of [tex]\( FePO_4 \)[/tex] is given as 150.82 g/mol.

2. Find the number of moles of [tex]\( FePO_4 \)[/tex]:
[tex]\[ \text{Number of moles of } FePO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{20.00 \text{ g}}{150.82 \text{ g/mol}} \approx 0.1326 \text{ moles} \][/tex]

3. Use the stoichiometry of the balanced equation:
According to the balanced chemical equation
[tex]\[ 2 FePO_4 + 3 Na_2SO_4 \rightarrow Fe_2(SO_4)_3 + 2 Na_3PO_4 \][/tex]
2 moles of [tex]\( FePO_4 \)[/tex] produce 1 mole of [tex]\( Fe_2(SO_4)_3 \)[/tex].

4. Calculate the moles of [tex]\( Fe_2(SO_4)_3 \)[/tex] produced:
[tex]\[ \text{Moles of } Fe_2(SO_4)_3 = \frac{\text{moles of } FePO_4}{2} = \frac{0.1326 \text{ moles}}{2} \approx 0.0663 \text{ moles} \][/tex]

5. Calculate the molar mass of [tex]\( Fe_2(SO_4)_3 \)[/tex]:
The molar mass of [tex]\( Fe_2(SO_4)_3 \)[/tex] is given as 399.88 g/mol.

6. Calculate the theoretical yield of [tex]\( Fe_2(SO_4)_3 \)[/tex]:
[tex]\[ \text{Theoretical yield} = \text{moles of } Fe_2(SO_4)_3 \times \text{molar mass of } Fe_2(SO_4)_3 = 0.0663 \text{ moles} \times 399.88 \text{ g/mol} \approx 26.51 \text{ g} \][/tex]

Therefore, 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 [tex]\( 26.51 \)[/tex] grams.

Among the given choices:
- 26.52 g (closest to the calculated theoretical yield)
- 53.04 g
- 150.8 g
- 399.9 g

The correct answer is: 26.52 g.
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