\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_K=x\\n_{Al}=y\end{matrix}\right.\) ( mol )
\(2K+2H_2O\rightarrow2KOH+H_2\)
x x 0,5x ( mol )
`@`TH1: Al chỉ tan một phần
\(n_{Al\left(OH\right)_3}=\dfrac{15,6}{78}=0,2\left(mol\right)\)
\(2Al+2KOH+2H_2O\rightarrow2KAlO_2+3H_2\)
x x x 1,5x ( mol )
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(KAlO_2+HCl+H_2O\rightarrow Al\left(OH\right)_3\downarrow+KCl\)
0,2 0,2 0,2 ( mol )
\(\Rightarrow n_{H_2}=1,5x+0,5x=1,5.0,2+0,5.0,2=0,4\left(mol\right)\)
`=>` không thỏa mãn đề bài
`@` TH2: Al tan hết
\(2Al+2KOH+2H_2O\rightarrow2KAlO_2+3H_2\)
y y 1,5y ( mol )
\(KAlO_2+HCl+H_2O\rightarrow Al\left(OH\right)_3\downarrow+KCl\)
0,2 0,2 0,2 ( mol )
\(\Rightarrow y=0,2\)
\(n_{H_2}=0,5x+1,5y=0,5\)
\(\Rightarrow x=0,4\)
\(\left\{{}\begin{matrix}m_K=0,4.39=15,6\left(g\right)\\m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)
\(V_{HCl}=\dfrac{0,2}{1}=0,2\left(M\right)\)
