Đáp án:
\(\begin{array}{l}
a)\\
{V_{{O_2}}} = 0,747l\\
b)\\
{m_{F{e_3}{O_4}}} = 3,87g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3Fe + 2{O_2} \to F{e_3}{O_4}\\
a)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{2,8}}{{56}} = 0,05mol\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} = \dfrac{1}{{30}}mol\\
{V_{{O_2}}} = n \times 22,4 = \dfrac{1}{{30}} \times 22,4 = 0,747l\\
b)\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{Fe}}}}{3} = \dfrac{{0,05}}{3} = \dfrac{1}{{60}}mol\\
{m_{F{e_3}{O_4}}} = n \times M = \frac{1}{{60}} \times 232 = 3,87g
\end{array}\)
