Đáp án:

\(\begin{array}{l}
b)\\
m = 36,45g\\
c)\\
{V_1} = 250
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
MgC{O_3} + 2HCl \to MgC{l_2} + C{O_2} + {H_2}O\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{MgC{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{m_{CaO}} = 23,82 - 0,15 \times 84 = 11,22g\\
{n_{CaO}} = \dfrac{{11,22}}{{56}} \approx 0,2\,mol\\
{n_{CaC{l_2}}} = {n_{CaO}} = 0,2\,mol\\
{n_{MgC{l_2}}} = {n_{MgC{O_3}}} = 0,15\,mol\\
m = 0,2 \times 111 + 0,15 \times 95 = 36,45g\\
c)\\
{n_{N{a_2}C{O_3}}} = \dfrac{{10,6}}{{106}} = 0,1\,mol\\
BTNT\,C:{n_{NaHC{O_3}}} = 0,15 - 0,1 = 0,05\,mol\\
BTNT\,Na:{n_{NaOH}} = 0,1 \times 2 + 0,05 = 0,25\,mol\\
{V_{{\rm{dd}}NaOH}} = \dfrac{{0,25}}{1} = 0,25l = 250ml \Rightarrow {V_1} = 250
\end{array}\)