The theoretical yield of Ca₃(PO₄)₂ is 372 g.
<h3>Calculation:</h3>
It is based on limiting reactants. The limiting reactant is the one that yields the fewest moles in the final product. The reaction ends when it runs out.
A balanced equation is given as:
3Ca(NO₃)₂ + 2Li₃PO₄ → 6LiNO₃ + Ca₃(PO₄)₂
Multiply the moles of each reactant by the mole ratio between it and Ca₃(PO₄)₂ in the balanced equation, so that the moles of the reactant cancel, leaving moles of Ca₃(PO₄)₂.
3.40 mol Ca(NO₃)₂ × 1 mol Ca₃(PO₄)₂/3 mol Ca(NO₃)₂ = 1.13 mol Ca₃(PO₄)₂
2.40 mol Li3PO4 × 1 mol Ca₃(PO₄)₂/2 mol Li₃PO₄ = 1.20 mol Ca₃(PO₄)₂
Ca(NO₃)₂ is the limiting reactant.
Now, calculate the mass of 1.20 mol Ca₃(PO₄)₂.
Molar mass Ca₃(PO₄)₂ = 310.174 g/mol
1 mol Ca₃(PO₄)₂ = 310.174 g/mol
1.20 mol Ca₃(PO₄)₂ = 1.20 mol × 310.174 g/mol
= 372 g
Hence, the theoretical yield of Ca₃(PO₄)₂ is 372 g.
Learn more about limiting reactants here:
brainly.com/question/28170241
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