Question

What is the weight of $CaO$ required to remove hardness of ${10}^{6}L$ of water containing $1.62gofCa(HC{O}_3{)}_2$ in $1L$ ?
The reaction is ,
$CaO+Ca(HC{O}_3{)}_2\to 2CaC{O}_3+H_{2}O$

• A. $9.5\times {10}^{3}g$
• B. $3.2\times {10}^{4}g$
• C. $2.3\times {10}^{6}g$
• D. $5.6\times {10}^{5}g$

Answer 1

The correct option is D $5.6\times {10}^{5}g$

Volume of water = ${10}^{6}L$

Weight of $Ca(HC{O}_3{)}_2=1.62g$

The reaction equation is written as,
$CaO+Ca(HC{O}_3{)}_2\to 2CaC{O}_3+H_{2}O$

$(n_f{)}_{Ca(HC{O}_3{)}_2}=2$

$\text{Molecular mass of}Ca(HC{O}_3{)}_2=40+2\times (1+12+48)=162gmo{l}^{-1}$
$\text{Equivalent mass}=\frac{\text{Molecular Weight}}{\text{n-factor}}=\frac{162}{2}$

Equivalents of $Ca(HC{O}_3{)}_2$ present in hard water is
$\frac{\text{Given Mass}}{\text{Equivalent mass}}$
$=\frac{1.62}{162/2}=0.02$

So the eq of $CaO$ required to remove hardness of $1L{H}_{2}O$ is 0.02.
Eq of $CaO$ required to remove hardness of ${10}^{6}Lis0.02\times {10}^6=2\times {10}^4$
$\therefore$ Mass of $CaO=\frac{2\times {10}^4\times 56}{2}=5.6\times {10}^{5}g$