Đáp án:
\(\begin{array}{l}
b)\\
{m_{F{e_3}{O_4}}} = 32,04g\\
c)\\
{V_{kk}} = 30,93l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
b)\\
{n_{Fe}} = \dfrac{{23,2}}{{56}} = \dfrac{{29}}{{70}}\,mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{29}}{{70}} \times \dfrac{1}{3} = \dfrac{{29}}{{210}}\,mol\\
{m_{F{e_3}{O_4}}} = \dfrac{{29}}{{210}} \times 232 \approx 32,04g\\
c)\\
{n_{{O_2}}} = \dfrac{{29}}{{70}} \times \dfrac{2}{3} = \dfrac{{29}}{{105}}\,mol\\
{V_{kk}} = \dfrac{{29}}{{105}} \times 5 \times 22,4 \approx 30,93l
\end{array}\)
