By decomposition of $50 g C a C O_{3}$ , $11 g$ of $C O_{2}$ is obtained then find % purity of $C a C O_{3}$ .

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Solution

$C a C O_{3} ⟶ C a O + C O_{2}$
Moles of $C O_{2} = \frac{11}{44} = 0.25$
Moles of $C a C O_{3} = 0.25$ (stoichiometry ratio $1 : 1$ )
Mass $= 100 \times 0.25 = 25$ $g m$ of pure $C a C O_{3}$
$% p u r i t y = \frac{25}{50} \times 100 = 50 %$

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