\(PTHH:FeS+2HCl\rightarrow FeCl_2+H_2S\)
Ta có:
\(n_{FeS}=\frac{5}{44}=0,11\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=18,26\left(g\right);m_{HCl}=3,625\left(g\right)\)
\(\Rightarrow n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow n_{NaOH}=0,05\left(mol\right)\)
\(T=\frac{0,05}{0,05}=1\)
\(Na_2S+H_2S\rightarrow NaHS+H_2O\)
Dung dịch A chỉ có : NaHS