Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\\
M + {H_2}S{O_4} \to MS{O_4} + {H_2}\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{{19,6}}{{98}} = 0,2\,mol\\
{n_M} = {n_{{H_2}S{O_4}}} = 0,2\,mol\\
{M_M} = \dfrac{{11,2}}{{0,2}} = 56g/mol \Rightarrow M:Fe\\
c)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
d)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{16}}{{160}} = 0,1\,mol\\
\text{ Ta có }:\dfrac{{{n_{{H_2}}}}}{3} < \dfrac{{{n_{F{e_2}{O_3}}}}}{1} \Rightarrow \text{ $Fe_2O_3$ dư }\\
{n_{F{e_2}{O_3}}} \text{ dư }= 0,1 - \dfrac{{0,2}}{3} = \dfrac{1}{{30}}\,mol\\
{n_{Fe}} = \dfrac{{0,2 \times 2}}{3} = \dfrac{2}{{15}}\,mol\\
m = \dfrac{1}{{30}} \times 160 + \dfrac{2}{{15}} \times 56 = 12,8g\\
2)\\
a)\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
b)\\
{n_{Cu}} = \dfrac{{12,8}}{{64}} = 0,2mol\\
{n_{{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
\text{ Ta có }:\dfrac{{{n_{Cu}}}}{2} < \dfrac{{{n_{{O_2}}}}}{1} \Rightarrow \text{ $O_2$ dư }\\
{n_{{O_2}}} = 0,3 - \dfrac{{0,2}}{2} = 0,2\,mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{V_{kk}} = (6,72 - 4,48) \times 5 = 11,2l
\end{array}\)
