Em tham khảo nha:
\(\begin{array}{l}
1)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
{K_2}O + {H_2}O \to 2KOH\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
{n_K} = 2{n_{{H_2}}} = 0,1\,mol\\
\% {m_K} = \dfrac{{0,1 \times 39}}{{12,4}} \times 100\% = 31,45\% \\
\% {m_{{K_2}O}} = 100 - 31,45 = 68,55\% \\
2)\\
a)\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{61,2}}{{102}} = 0,6\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{9,8 \times 30\% }}{{98}} = 0,03\,mol\\
{n_{A{l_2}{O_3}}} > \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow A{l_2}{O_3} \text{ dư , tính theo $H_2SO_4$ }\\
{n_{A{l_2}{O_3}}} \text{ phản ứng } = \dfrac{{0,03}}{3} = 0,01\,mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,03}}{3} = 0,01\,mol\\
{m_{{\rm{dd}}spu}} = 0,01 \times 102 + 9,8 = 10,82g\\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,01 \times 342}}{{10,82}} \times 100\% = 31,61\%
\end{array}\)
