Em tham khảo nha :
\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:N{a_2}C{O_3}(a\,mol),CaC{O_3}(b\,mol)\\
a + b = 0,3(1)\\
106a + 100b = 30,6(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,2\\
{m_{N{a_2}C{O_3}}} = 0,1 \times 106 = 10,6g\\
{m_{CaC{O_3}}} = 30,6 - 10,6 = 20g\\
b)\\
{n_{HCl}} = 2{n_{C{O_2}}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
{m_{ddHCl}} = \dfrac{{21,9 \times 100}}{{20}} = 109,5g\\
c)\\
{m_{ddspu}} = 30,6 + 109,5 - 0,3 \times 44 = 126,9g\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,2mol\\
{m_{NaCl}} = 0,2 \times 58,5 = 11,7g\\
{n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{m_{CaC{l_2}}} = 0,2 \times 111 = 22,2g\\
C{\% _{NaCl}} = \dfrac{{11,7}}{{126,9}} \times 100\% = 9,22\% \\
C{\% _{CaC{l_2}}} = \dfrac{{22,2}}{{126,9}} \times 100\% = 17,5\%
\end{array}\)
