Đáp án:

5, \({m_{FeS{O_4}}} = 30,4g\) và \({V_{{H_2}}} = 4,48l\)

6, \({V_{C{l_2}}} = 6,496l\)

Giải thích các bước giải:

5,

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,2mol\\
 \to {n_{FeS{O_4}}} = {n_{Fe}} = 0,2mol\\
 \to {m_{FeS{O_4}}} = 30,4g\\
 \to {n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
 \to {V_{{H_2}}} = 4,48l
\end{array}\)

6,

\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{m_{{O_2}}} = 22,12 - 21,16 = 0,96g\\
 \to {n_{{O_2}}} = 0,03mol\\
 \to {n_{{K_2}Mn{O_4}}} = {n_{Mn{O_2}}} = {n_{{O_2}}} = 0,03mol\\
 \to {m_{KMn{O_4}}}dư= 21,16 - 0,03 \times (197 + 87) = 12,64g\\
 \to {n_{KMn{O_4}}}dư= 0,08mol\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{K_2}Mn{O_4} + 8HCl \to 2KCl + MnC{l_2} + 2C{l_2} + 4{H_2}O\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}}}dư+ 2{n_{{K_2}Mn{O_4}}} + {n_{Mn{O_2}}} = 0,29mol\\
 \to {V_{C{l_2}}} = 6,496l
\end{array}\)