Sau phản ứng nung, thu được Y gồm Al, S, Al2S3
\(m_X=m_S=5,76\left(g\right)\)
\(\Rightarrow n_{S_{dư}}=0,18\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2S_3+6HCl\rightarrow2AlCl_3+3H_2S\)
\(n_T=n_{H2}+n_{H2S}=0,36\left(mol\right)\)
Gọi a là mol H2, b là mol H2S
\(\Rightarrow a+b=0,36\left(1\right)\)
\(\overline{M_T}=13.2=26\)
\(\Rightarrow\frac{2a+34b}{a+b}=26\Rightarrow24a-8b=0\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,09\\b=0,27\end{matrix}\right.\)
\(n_{H2S}=0,27\left(mol\right)\Rightarrow n_{Al2S3}=0,09\left(mol\right)\)
\(2Al+3S\underrightarrow{^{to}}Al_2S_3\)
\(\Rightarrow n_{S_{pư}}=0,27\left(mol\right)\)
Ban đầu có 0,27 + 0,18 = 0,45 mol S
\(\Rightarrow H=\frac{0,27.100}{0,45}.100\%=60\%\)
