Đáp án:

\(\begin{array}{l}
a)\\
\% CaC{O_3} = 33,3\% \\
\% N{a_2}S{O_4} = 66,7\% \\
b)\\
{V_{NaOH}} = 3,2M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,1mol\\
{m_{CaC{O_3}}} = 0,1 \times 100 = 10g\\
\% CaC{O_3} = \dfrac{{10}}{{30}} \times 100\%  = 33,3\% \\
\% N{a_2}S{O_4} = 100 - 33,3 = 66,7\% \\
b)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
C{O_2} + NaOH \to NaHC{O_3}\\
hh:N{a_2}C{O_3}(a\,mol),NaHC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
2a - 3b = 0
\end{array} \right.\\
 \Rightarrow a = 0,06;b = 0,04\\
{n_{NaOH}} = 2{n_{N{a_2}C{O_3}}} + {n_{NaHC{O_3}}} = 0,16mol\\
{V_{NaOH}} = \dfrac{{0,16}}{{0,05}} = 3,2M
\end{array}\)