a,
Dư Fe => S hết
\(n_{khi}=0,6\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(FeS+H_2SO_4\rightarrow FeSO_4+H_2S\)
Gọi a là mol Fe dư; b là mol FeS
\(\Rightarrow a+b=0,6\left(1\right)\)
\(Fe+S\underrightarrow{^{to}}FeS\)
\(n_{Fe\left(pư\right)}=n_S=n_{FeS}=b\left(mol\right)\)
\(\Rightarrow56\left(a+b\right)+32b=46,4\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\)
X có 0,2+0,4= 0,6 mol Fe
\(\%_{Fe}=\frac{0,6.56.100}{46,4}=72,41\%\)
\(\%_S=100\%-72,41\%=27,59\%\)
b,\(n_{H2S}=0,4\left(mol\right)\)
\(CuSO_4+H_2S\rightarrow CuS+H_2SO_4\)
\(\Rightarrow n_{CuS}=0,4\left(mol\right)\)
\(\Leftrightarrow m=38,4\left(g\right)\)
