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Sagot :
Percentage yield tells the percentage ratio of actual to theoretical yield. The theoretical yield of iron(III) sulfate is 26.52 g.
What is a theoretical yield?
The theoretical yield is the product formation from the complete modification of the limiting reagent in a chemical reaction.
The balanced chemical reaction is shown as,
[tex]\rm 2FePO_{4} + 3Na_{2}SO_{4} \rightarrow Fe_{2}(SO_{4})_{3} + 2Na_{3}PO_{4}[/tex]
Moles of ferric phosphate is calculated as:
[tex]\begin{aligned} \rm n &= \rm \dfrac{mass}{molar \; mass}\\\\&= \dfrac{20}{150.82}\\\\&= 0.1326\;\rm mol\end{aligned}[/tex]
From the reaction,
[tex]\dfrac{1 \;\rm mol \; Fe_{2}(SO_{4})_{3} }{2 \;\rm mol \; FePO_{4}}&= \dfrac{\rm X}{ 0.1326 \;\rm mol \; FePO_{4}}[/tex]
Solving for X:
[tex]\begin{aligned} \rm X &= (\dfrac{0.1326}{2}) \times 1 \rm \; mol\\\\&= 0.0663 \;\rm mol \;Fe_{2}(SO_{4})_{3}\end{aligned}[/tex]
Calculating mass from moles:
[tex]\begin{aligned}\rm mass &= \rm moles \times molar \rm \; mass\\\\&= 0.0663 \;\rm mol \times 399.88 \;\rm g/mol \\\\&= 26.51\;\rm g\end{aligned}[/tex]
Therefore, option A. 26.52 gm is the mass of iron(III) sulfate.
Learn more about theoretical yield here:
https://brainly.com/question/8575227
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