Percentage yield tells the percentage ratio of actual to theoretical yield. The theoretical yield of iron(III) sulfate is 26.52 g.
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.
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