When 20   g of CaCO 3 is treated with 20   g HCl , the mass of CO 2 formed would be

So according to the question when the reaction takes place

$\begin{array}{ccccccccc}{\mathrm{CaCO}}_3& +& 2\mathrm{HCl}& \to & {\mathrm{CaCl}}_2& +& {\mathrm{H}}_2\mathrm{O}& +& {\mathrm{CO}}_2\\ 1\mathrm{mol}& & 2 \mathrm{mol}& & & & & & 1 \mathrm{mol}\\ 100 \mathrm{g}& & 73 \mathrm{g}& & & & & & 44 \mathrm{g}\end{array}$

Let ${\mathrm{CaCO}}_3$ be completely consumed in the reaction

So that 

$100 \mathrm{g}$ of ${\mathrm{CaCO}}_3$ give $44 \mathrm{g}$ of ${\mathrm{CO}}_2$

Thus,$20 \mathrm{g}$ of ${\mathrm{CaCO}}_3$ give

$\frac{44}{100} \times 20 \mathrm{g} = 8.8 \mathrm{g} {\mathrm{CO}}_2$