Đáp án:

\(\begin{array}{l}
{m_{Fe}} = 5,6g\\
C{M_{{H_2}S{O_4}}} = 1M
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}(1)\\
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl(2)\\
FeS{O_4} + BaC{l_2} \to BaS{O_4} + FeC{l_2}(3)
\end{array}\)

\(\begin{array}{l}
a.\\
{n_{{H_2}}} = 0,1mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
 \to {m_{Fe}} = 5,6g
\end{array}\)

\(\begin{array}{l}
b.\\
{n_{{\rm{BaS}}{O_4}}} = 0,2mol\\
 \to {n_{{\rm{BaS}}{O_4}(3)}} = {n_{FeS{O_4}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}(dư)}} = {n_{{\rm{BaS}}{O_4}(2)}} = 0,2 - 0,1 = 0,1mol
\end{array}\)

\(\begin{array}{l}
{n_{{H_2}S{O_4}(1)}} = {n_{{H_2}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}(banđầu)}} = 0,2mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,2}}{{0,2}} = 1M
\end{array}\)