Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_K} = 2{n_{{H_2}}} = 0,2mol\\
{m_K} = 0,2 \times 39 = 7,8g\\
{m_{{H_2}O}} = {m_{{\rm{dd}}}} - {m_K} + {m_{{H_2}}} = 200 - 7,8 + 0,1 \times 2 = 192,4g\\
b)\\
{n_{KOH}} = 2{n_{{H_2}}} = 0,2mol\\
{m_{KOH}} = 0,2 \times 56 = 11,2g\\
C{\% _{KOH}} = \dfrac{{11,2}}{{200}} \times 100\%  = 5,6\% \\
2)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{0,81}}{{27}} = 0,03mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{98 \times 20}}{{100}} = 19,6g\\
{n_{{H_2}S{O_4}}} = \dfrac{{19,6}}{{98}} = 0,2mol\\
\dfrac{{0,03}}{2} < \dfrac{{0,2}}{3} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - \frac{3}{2}{n_{Al}} = 0,155mol\\
{m_{{H_2}S{O_4}d}} = 0,155 \times 98 = 15,19g\\
b)\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,045mol\\
{m_{{\rm{dd}}spu}} = 0,81 + 98 - 0,045 \times 2 = 98,72g\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,015mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,015 \times 342 = 5,13g\\
C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{5,13}}{{98,72}} \times 100\%  = 5,2\% \\
C{\% _{{H_2}S{O_4}d}} = \dfrac{{15,19}}{{98,72}} \times 100\%  = 15,4\% 
\end{array}\)