Đáp án:

\(\begin{array}{l}
a)\\
{V_{{O_2}}} = 5,6l\\
b)\\
{n_{{H_2}S{O_4}}} = 0,5\,mol\\
c)\\
{V_{{\rm{dd}}HCl}} = 1l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
a)\\
4Fe + 3{O_2} \xrightarrow{t^0}  2F{e_2}{O_3}\\
2Fe + {O_2} \xrightarrow{t^0} 2FeO\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
BTKL:{m_{kl}} + {m_{{O_2}}} = {m_{hh}} \Rightarrow {m_{{O_2}}} = 31,2 - 23,2 = 8g\\
{n_{{O_2}}} = \dfrac{8}{{32}} = 0,25\,mol\\
{V_{{O_2}}} = 0,25 \times 22,4 = 5,6l\\
b)\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_O} = 2{n_{{O_2}}} = 0,5\,mol\\
{n_{{H_2}S{O_4}}} = {n_O} = 0,5\,mol\\
c)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{HCl}} = 2{n_O} = 1\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{1}{1} = 1l
\end{array}\)