Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 16,8g\\
{m_{Zn}} = 13g\\
c)\\
{C_M}{H_2}S{O_4} = 2M\\
d)\\
{m_1} = 27g\\
{m_2} = 24g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
b)\\
hh:Fe(a\,mol),Zn(b\,mol)\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
\left\{ \begin{array}{l}
56a + 65b = 29,8\\
a + b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,2\\
{m_{Fe}} = 0,3 \times 56 = 16,8g\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,5\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,5}}{{0,25}} = 2M\\
d)\\
FeS{O_4} + 2NaOH \to Fe{(OH)_2} + N{a_2}S{O_4}\\
ZnS{O_4} + 2NaOH \to Zn{(OH)_2} + N{a_2}S{O_4}\\
Zn{(OH)_2} + 2NaOH \to N{a_2}Zn{O_2} + 2{H_2}O\\
{n_{Fe{{(OH)}_2}}} = {n_{FeS{O_4}}} = 0,3\,mol\\
{m_1} = 0,3 \times 90 = 27g\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \xrightarrow{t^0} 4Fe{(OH)_3}\\
2Fe{(OH)_3} \xrightarrow{t^0} F{e_2}{O_3} + 3{H_2}O\\
{n_{Fe{{(OH)}_3}}} = {n_{Fe{{(OH)}_2}}} = 0,3\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{m_2} = 0,15 \times 160 = 24g
\end{array}\)