Phản ứng xảy ra:
\(2Al+3S\underrightarrow{^{t^o}}Al_2S_3\)
Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_S:y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow27x+32y=10,2\left(g\right)\)
Vì cho Y tác dụng với HCl thu được hỗn hợp khí nên Al dư
\(\Rightarrow n_{Al2S3}=\frac{1}{3}n_S=\frac{y}{3}\left(mol\right)\)
\(\Rightarrow n_{Al\left(dư\right)}=x-\frac{2y}{3}\left(mol\right)\)
\(Al_2S_3+6HCl\rightarrow2AlCl_3+3H_2S\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\Rightarrow n_{H2S}=3n_{Al2S3}=y\left(mol\right);n_{H2}=\frac{3}{2}n_{Al}=1,5x-y\left(mol\right)\)
Ta có:
\(M_Z=9.2=18\)
Áp dụng quy tắc đường chéo:
\(\Rightarrow\frac{n_{H2S}}{n_{H2}}=\frac{y}{1,5x-y}=\frac{16}{16}\Rightarrow1,5x-y=y\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
\(\Rightarrow m=0,2.27=5,4\left(g\right)\Rightarrow\%m_{Al}=\frac{5,4}{10,2}.100\%=52,94\%\)
