Question

Temporary hardness is due to $HC{O}_3^{-}$ of $M{g}^{2+}$ and $C{a}^{2+}$. It is removed by addition of $CaO$.
$Ca(HC{O}_3{)}_2+CaO\to 2CaC{O}_3+H_{2}O$
Mass of $CaO$ (in g) required to produce $2$ g of $CaC{O}_3$ precipitate is:

• A. $2.00$
• B. $0.56$
• C. $0.28$
• D. $1.12$

Answer 1

Answer: (B)

$0.56$

Molecular weight of $CaC{O}_3$ and $CaO$ are $100$ g/mol and $56$ g/mol respectively.
$2$ equivalents of $CaC{O}_3$ requires $1$ equivalents of $CaO$.

Amount of $CaO$ required to produce $2$ g of $CaC{O}_3$ $=\frac{2\times 56}{2\times 100}=0.56$