Calculate the mass of CaO required to remove the hardness of ${10}^6$ litre of water containing 1.62 gm of calcium bicarbonate per litre

$CaO+Ca{\left(HC{O}_3\right)}_2\underset{ }{\overset{ }{\to }}2CaC{O}_3+H_{2}O$

$\therefore$ Equivalent Wt. of $Ca{\left(HC{O}_3\right)}_2$ present in hard water (1 litre) $\frac{1.62}{162/2}=0.02$

$\therefore$Equivalent Wt. of  CaO required to remove$Ca{\left(HC{O}_3\right)}_2$ in 1 litre = 0.02

$\therefore$Equivalent Wt. of CaO required to remove $Ca{\left(HC{O}_3\right)}_2$ in ${10}^6$ litre$=0.02\times {10}^6$

$\therefore$mass of $CaO=0.02\times {10}^6\times \frac{56}{2}g=5.6\times {10}^{5}gm$