Giải thích các bước giải:
\(\begin{array}{l}
5.\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = 0,1mol\\
\to {n_{CaC{l_2}}} = {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,1mol\\
\to {m_{CaC{l_2}}} = 0,1 \times 111 = 11,1g\\
{m_{{\rm{dd}}HCl}} = 250 \times 1,2 = 300g\\
\to {m_{{\rm{dd}}CaC{l_2}}} = {m_{CaC{O_3}}} + {m_{{\rm{dd}}HCl}} - {m_{C{O_2}}} = 10 + 300 - 0,1 \times 44 = 305,6g\\
\to C{\% _{{\rm{dd}}CaC{l_2}}} = \dfrac{{11,1}}{{305,6}} \times 100\% = 3,63\% \\
7.\\
X + 2HCl \to XC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,3mol\\
\to {n_X} = {n_{{H_2}}} = 0,3mol\\
\to {M_X} = \dfrac{{7,2}}{{0,3}} = 24\\
\to X = Mg\\
\end{array}\)