Em tham khảo nha:

\(\begin{array}{l}
7)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Al}} = 0,3 \times \frac{2}{3} = 0,2\,mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,3}}{{0,3}} = 1M\\
10)\\
a)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07mol\\
{n_{NaOH}} = \dfrac{{6,4}}{{40}} = 0,16\,mol\\
\text{ Lập tỉ lệ }:\dfrac{{{n_{C{O_2}}}}}{1} < \dfrac{{{n_{NaOH}}}}{2}(0,07 < \dfrac{{0,16}}{2})\\
 \Rightarrow \text{ NaOH dư tính theo $CO_2$} \\
{n_{NaOH}} \text{ dư}= 0,16 - 0,07 \times 2 = 0,02\,mol\\
{m_{NaOH}} \text{ dư} = 0,02 \times 40 = 0,8g\\
c)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,07\,mol\\
{m_{N{a_2}C{O_3}}} = 0,07 \times 106 = 7,42g
\end{array}\)