Question

A 0.700 g unknown mixture with CaCl2 reacts with aqueous sodium phosphate to give 0.425 g Ca3(PO4)2. Show calculation for the percentage CaCl2?

Answer 1

The percentage CaCl₂ : 65.14%

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

3CaCl₂+2Na₃PO₄⇒Ca₃(PO₄)₂ + 6NaCl

mol Ca₃(PO₄)₂(MW=310.18 g/mol) :

$\tt \dfrac{0.425}{310.18}=0.00137$

mol ratio CaCl₂ : Ca₃(PO₄)₂ = 3 : 1, so mol CaCl₂ :

$\tt \dfrac{3}{1}\times 0.00137=0.00411$

mass CaCl₂ (MW=110.98 g/mol) :

$\tt 0.00411\times 110.98=0.456~g$

The percentage CaCl₂ :

$\tt \%CaCl_2~in~mixture=\dfrac{0.456}{0.7}\times 100\%=65.14\%$