Đáp án:

\(\begin{array}{l}
a)\\
{m_K} = 15,6g\\
{m_{Al}} = 8,1g\\
b)\\
{V_{C{l_2}}} = 11,2l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
4K + {O_2} \xrightarrow{t^0} 2{K_2}O\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
hh:K(\,a\,mol),Al(b\,mol)\\
{n_{{K_2}O}} = \dfrac{{{n_K}}}{2} = \dfrac{a}{2}mol\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = \dfrac{b}{2}mol\\
39a + 27b = 23,7(1)\\
47a + 51b = 34,1(2)\\
 \text{Từ (1) và (2)}\Rightarrow a = 0,4;b = 0,3\\
{m_K} = n \times M = 0,4 \times 39 = 15,6g\\
{m_{Al}} = n \times M = 0,3 \times 27 = 8,1g\\
b)\\
2K + C{l_2} \to 2KCl\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
\dfrac{{{m_a}}}{{{m_b}}} = \dfrac{{23,7}}{{9,48}} = 2,5\\
 \Rightarrow {n_{K(b)}} = \dfrac{{0,4}}{{2,5}} = 0,16mol\\
{n_{Al(b)}} = \dfrac{{0,3}}{{2,5}} = 0,12mol\\
{n_{C{l_2}}} = 2{n_{K(b)}} + \frac{3}{2}{n_{Al(b)}} = 0,5mol\\
{V_{C{l_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l
\end{array}\)