a,
\(\left\{{}\begin{matrix}n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\\n_S=\frac{1,6}{32}=0,05\left(mol\right)\end{matrix}\right.\)
\(Fe+S\rightarrow FeS\)
Nên Fe dư 0,05 mol
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05__________0,05___________
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
0,05______________0,05_______
\(\Rightarrow\%V_{H2}=\%n_{H2}=\frac{0,05}{0,1}.100\%=50\%\)
\(\Rightarrow\%V_{H2S}=5n_{H2S}=50\%\)
b, \(n_{NaOH}=0,0125\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,0125___ 0,0125______________
\(n_{HCl}=0,1+0,1+0,125=0,2125\left(mol\right)\)
\(\Rightarrow CM_{HCl}=\frac{0,2125}{0,5}=0,425M\)