Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
b)\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{13}}{{65}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{24,5}}{{98}} = 0,25mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,25}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
{V_{{H_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{{H_2}S{O_4}d}} = 0,25 - 0,2 = 0,05mol\\
{m_{{H_2}S{O_4}d}} = n \times M = 0,05 \times 98 = 4,9g\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,2mol\\
{m_{ZnSO4}} = n \times M = 0,2 \times 161 = 32,2g\\
2)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{0,81}}{{27}} = 0,03mol\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{2,19}}{{36,5}} = 0,06mol\\
\dfrac{{0,03}}{2} > \dfrac{{0,06}}{6} \Rightarrow Al\text{ dư}\\
{n_{A{l_d}}} = 0,03 - \dfrac{{0,06}}{6} \times 2 = 0,01mol\\
{m_{Al}} = n \times M = 0,01 \times 27 = 0,27g\\
b)\\
{n_{AlC{l_3}}} = \dfrac{2}{6}{n_{HCl}} = 0,02mol\\
{m_{AlC{l_3}}} = n \times M = 0,02 \times 133,5 = 2,67g\\
3)\\
a)\\
4Na + {O_2} \to 2N{a_2}O\\
{n_{Na}} = \dfrac{m}{M} = \dfrac{{4,6}}{{23}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{0,448}}{{22,4}} = 0,02mol\\
\dfrac{{0,2}}{4} > \dfrac{{0,02}}{1} \Rightarrow Na\text{ dư}\\
{n_{N{a_d}}} = 0,2 - 4 \times 0,02 = 0,12mol\\
{m_{N{a_d}}} = n \times M = 0,12 \times 23 = 2,76g\\
b)\\
{n_{N{a_2}O}} = 2{n_{{O_2}}} = 0,04mol\\
{m_{N{a_2}O}} = n \times M = 0,04 \times 62 = 2,48g
\end{array}\)
