Em tham khảo nha :
\(\begin{array}{l}
3)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{21,6}}{{27}} = 0,8mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 1,2mol\\
{V_{{H_2}}} = 1,2 \times 22,4 = 26,88l\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 1,2mol\\
{m_{{H_2}S{O_4}}} = 1,2 \times 98 = 117,6g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{117,6 \times 100}}{{29,4}} = 400g\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \frac{{{n_{Al}}}}{2} = 0,4mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,4 \times 342 = 136,8g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{136,8}}{{21,6 + 400 - 1,2 \times 2}} \times 100\% = 32,633\% \\
4)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Na}} = 2{n_{{H_2}}} = 0,4mol\\
{m_{Na}} = 0,4 \times 23 = 9,2g\\
b)\\
{n_{NaOH}} = {n_{Na}} = 0,4mol\\
{m_{NaOH}} = 0,4 \times 40 = 16g\\
C{\% _{NaOH}} = \dfrac{{16}}{{9,2 + 191,2 - 0,2 \times 2}} \times 100\% = 8\%
\end{array}\)
