Đáp án:
\(\begin{array}{l}
b)\\
{V_{{O_2}}} = 4,48l\\
c)\\
{m_{cr}} = 20g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
b)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{16,8}}{{56}} = 0,3mol\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} = 0,2mol\\
{V_{{O_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
c)\\
F{e_3}{O_4} + 4{H_2} \xrightarrow{t^0} 3Fe + 4{H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{Fe}}}}{3} = 0,1mol\\
\dfrac{{0,2}}{4} < \dfrac{{0,1}}{1} \Rightarrow F{e_3}{O_4}\text{ dư}\\
{n_{F{e_3}{O_4}d}} = {n_{F{e_3}{O_4}}} - \dfrac{{{n_{{H_2}}}}}{4} = 0,05mol\\
{m_{F{e_3}{O_4}d}} = n \times M = 0,05 \times 232 = 11,6g\\
{n_{Fe}} = \frac{3}{4}{n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = n \times M = 0,15 \times 56 = 8,4g\\
{m_{cr}} = {m_{F{e_3}{O_4}d}} + {m_{Fe}} = 11,6 + 8,4 = 20g
\end{array}\)