When a sample of 100ML of 0.0580M stannous fluoride is mixed with 25mL of .100M Na3(PO4)2, .555g precipitate of stannous carbonate is obtained. Calculate the theoretical and percent yield of Sn3(PO4)2

Answer 1

stannous fluoride = SnF₂

sodium phosphate is Na₃PO₄ NOT Na3(PO4)2 as indicated in your question

3SnF₂ + 2Na₃PO₄ ==> Sn₃(PO₄)₂ + 6NaF ... balanced equation

moles SnF₂ present = 0.100 L x 0.0580 mol/L = 0.00580 mols

moles Na₃PO₄ present = 0.025 L x 0.100 mol/L = 0.0025 mols

The Na₃PO₄ is limiting as indicated by the 3:2 mol ratio of SnF₂ to Na₃PO₄

Theoretical yield (in g) of Sn₃(PO₄)₂:

0.0025 mols Na₃PO₄ x 1 mol Sn₃(PO₄)₂ / 2 mol Na₃PO₄ x 546 g Sn₃(PO₄)₂/mol = 0.683 g Sn₃(PO₄)₂

Percent yield = actual yield / theoretical yield (x100%)

% yield = 0.555 g / 0.683 g (x100%) = 81.3% yield