Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
hh:{H_2}(a\,mol),C{O_2}(b\,mol)\\
{M_{hh}} = 3,1{M_{{H_2}}} = 6,2\\
\left\{ \begin{array}{l}
a + b = 1,62\\
2a + 44b = 10,044
\end{array} \right.\\
\Rightarrow a = 1,458mol;b = 0,162mol\\
{n_{HCl}} = 2{n_{{H_2}}} + 2{n_{C{O_2}}} = 3,24mol\\
{V_{HCl}} = \dfrac{{3,24}}{{0,5}} = 6,48l\\
b)\\
{n_{Fe}} = {n_{{H_2}}} = 1,458mol\\
{m_{Fe}} = 1,458 \times 56 = 81,648g\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,162mol\\
{m_{CaC{O_3}}} = 0,162 \times 100 = 16,2g\\
\% Fe = \dfrac{{81,648}}{{81,648 + 16,2}} \times 100\% = 83,44\% \\
\% CaC{O_3} = 100 - 83,44 = 16,56\%
\end{array}\)