Đáp án:
a) 0,35l
b) 4,48l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{{\rm{dd}}NaOH}} = 50 \times 1,28 = 64g\\
{n_{NaOH}} = \dfrac{{64 \times 25\% }}{{40}} = 0,4\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = 0,2 + 0,15 = 0,35\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,35}}{1} = 0,35l\\
b)\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,15\\
56a + 24b = 6,8
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,05\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = 0,1 \times 1,5 + 0,05 = 0,2\,mol\\
{V_{S{O_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)
