when 10g of $CaC{O}_3$ is treated  with 10g of HCl, mass of $C{O}_2$  will produced is  $n\times {10}^{ – 1}.$Then the 
 value of n is

$$\begin{gathered} CaC{O}_3 + 2HCl \to CaC{l}_2 + H_{2}O + C{O}_2 \\ 100 73 44 \end{gathered}$$

$CaC{O}_3$ is limiting reagent

100g $CaC{O}_3$ → 44g of $C{O}_2$
10 g $CaC{O}_3$ → 
mass of $C{O}_2$ produced = $\frac{440}{100}$ =4.4 g