Question

Following reaction is used to remove temporary hardness which is due to dissolved bicarbonates of $C{a}^{2+}$ and $M{g}^{2+}$.

$Ca(HC{O}_3{)}_2+Ca(OH{)}_2\to 2CaC{O}_3\downarrow +2H_{2}O$

Actually lime ($CaO$) is added which in $H_{2}O$ forms $Ca(OH{)}_2$ and precipitates $CaC{O}_3$ $CaO+H_{2}O\to Ca(OH{)}_2$

If $10$ L of hard water requires $0.56$ g of $CaO$, temporary hardness of $CaC{O}_3$ is $x\times {10}^2$ ppm. Find $x$.

Answer 1

The molar mass of CaO is $56$ g/mol.

The number of moles in $0.56$ g CaO $=\frac{0.56}{56}=0.01$ moles.

$1$ mole of CaO reacts with $1$ mole of calcium hydroxide to give $2$ moles of $CaC{O}_3$.
$0.01$ moles of CaO will give $0.01\times 2=0.02$ moles of $CaC{O}_3$
Molar mass of $CaC{O}_3$ is $100$ g/mol

The mass of $CaC{O}_3$ $=0.02\times 100=2$ g.
$2$ g $CaC{O}_3$ corresponds to $10$ L of hard water.

Hence, temporary hardness of $CaC{O}_3$ is:
$=\frac{2g}{10L\times {10}^{3}mL/L}\times {10}^6=2\times {10}^2$ ppm which is equal to $2\times {10}^2$ ppm .

Hence, $x=2$