Question

if 49.0 g of H3PO4 is reacted with excess KOH, determine the theoretical yield of K3PO4 if you isolate 49.0g of K3PO4?

Answer 1

H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

M(H₃PO₄) = 98.0 g/mol

n(H₃PO₄) $= \frac{49.0}{98.0}= 0.5 \;mol$

According to the reaction:

n(H₃PO₄)=n(K₃PO₄) = 0.5 mol

M(K₃PO₄) = 212.0 g/mol

m(K₃PO₄) $= 0.5 \times 212.0 = 106.0 \;g$

Proportion:

106.0 g – 100 %

49.0 g – x

$x = \frac{49.0 \times 100}{106.0} = 46.22 \; \%$

Answer: 46.22 %