Question

A solution containing 9.15 g of calcium nitrate is added to a solution containing 9.15 g of potassium phosphate. What is the theoretical yield of the precipitate in grams? Report your answer rounded to the hundredths of a gram.

Answer 1

Molar mass of Ca(NO₃) ₂ = 40.08 + ( 2 $\times$ 14.0067) + ( 6 $\times$ 16.00) = 164.09 g / mol

Molar mass of Ca₃ (PO₄)₂ = ( 3 $\times$ 40.08 ) + ( 2 $\times$ 30.97) + ( 8 $\times$ 16.00) = 310.18 g/mol

Molar mass of K₃PO₄ = ( 3 $\times$ 39.10 ) + 30.97 + ( 4 $\times$ 16.00) = 212.27 g/ mol

Consider a precipitation reaction, 3 Ca(NO₃) ₂ (aq) + 2 K₃PO₄ (aq) $\rightarrow$ Ca₃ (PO₄)₂ (s) + 6 KNO ₃ (aq)

From reaction , 3 mol Ca(NO₃) ₂ $\equiv$ 2 mol K₃PO₄$\equiv$1 mol Ca₃ (PO₄)₂

3 $\times$ 164.09 g Ca(NO₃) ₂ $\equiv$ 2 $\times$ 212.97 g K₃PO₄

9.15 g  Ca(NO₃) ₂ $\equiv$ 2 $\times$ 212.97 $\times$ 9.15 / 3 $\times$ 164.09 g K₃PO₄

9.15 g  Ca(NO₃) ₂ $\equiv$ 7.917 g K₃PO₄

i e 9.15 g calcium nitrate requires 7.917 g potassium phosphate. Hence, potassium phosphate is excess reactant because its amount (9.15 g) is excess than required amount ( 7.917 g)

Calcium nitrate is limiting reactant , hence mass of product will depend on mass of calcium nitrate.

We have relation,  3 mol Ca(NO₃) ₂ $\equiv$ 1 mol Ca₃ (PO₄)₂

3 $\times$ 164.09 g Ca(NO₃) ₂ $\equiv$ 310.18 g Ca₃ (PO₄)₂

9.15 g  Ca(NO₃) ₂ $\equiv$  310.18 $\times$ 9.15 / ( 3 $\times$ 164.09) g Ca₃ (PO₄)₂

9.15 g  Ca(NO₃) ₂ $\equiv$ 5.76 g Ca₃ (PO₄)₂

ANSWER : Theoretical yield of precipitate in grams = 5.76 g