Em tham khảo nha:

\(\begin{array}{l}
10)\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
{n_{AlC{l_3}}} = \dfrac{{26,7}}{{133,5}} = 0,2\,mol\\
{n_{C{l_2}}} = 0,2 \times \frac{3}{2} = 0,3\,mol\\
{m_{C{l_2}}} = 0,3 \times 71 = 21,3g\\
12)\\
{V_{C{l_2}}} = 0,3 \times 22,4 = 6,72l\\
13)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
{n_{C{l_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Fe}} = 0,3 \times \dfrac{2}{3} = 0,2\,mol\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
{n_{FeC{l_3}}} = {n_{Fe}} = 0,2\,mol\\
{m_{FeC{l_3}}} = 0,2 \times 162,5 = 32,5g\\
14)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{Mn{O_2}}} = \dfrac{{10,44}}{{87}} = 0,12\,mol\\
{n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,12\,mol\\
{V_{C{l_2}}} = 0,12 \times 22,4 = 2,688l
\end{array}\)