\[
3\text{Fe} + 2\text{O}_2 \to \text{Fe}_3\text{O}_4
\]

\[
n_{\text{Fe}} = \frac{m_{\text{Fe}}}{M_{\text{Fe}}} = \frac{16,8}{56} = 0,3 \, \text{mol}
\]

\[
n_{\text{O}_2} = \frac{V}{22,4} = \frac{7,437}{22,4} = 0,332 \, \text{mol}
\]

\[
3 \text{ mol Fe} : 2 \text{ mol O}_2
\]

\[
0,332 \times \frac{3}{2} = 0,498 \, \text{mol}
\]

\[
n_{\text{O}_2 \, \text{dư}} = n_{\text{O}_2} - \frac{n_{\text{Fe}} \cdot 2}{3} = 0,332 - \frac{0,3 \cdot 2}{3} = 0,132 \, \text{mol}
\]

\[
m_{\text{O}_2 \, \text{dư}} = n_{\text{O}_2 \, \text{dư}} \cdot M_{\text{O}_2} = 0,132 \cdot 32 = 4,224 \, \text{gam}
\]

\[
V_{\text{O}_2 \, \text{dư}} = n_{\text{O}_2 \, \text{dư}} \cdot 22,4 = 0,132 \cdot 22,4 = 2,957 \, \text{lít}
\]

\[
n_{\text{Fe}_3\text{O}_4} = \frac{n_{\text{Fe}}}{3} = \frac{0,3}{3} = 0,1 \, \text{mol}
\]

\[
m_{\text{Fe}_3\text{O}_4} = n_{\text{Fe}_3\text{O}_4} \cdot M_{\text{Fe}_3\text{O}_4} = 0,1 \cdot 232 = 23,2 \, \text{gam}
\]

Câu 5: Tính số mol các chất sau:

\[
m_{\text{CuSO}_4} = \frac{3,2}{100} \cdot 100 = 3,2 \, \text{gam}
\]
\[
n_{\text{CuSO}_4} = \frac{m_{\text{CuSO}_4}}{M_{\text{CuSO}_4}} = \frac{3,2}{160} = 0,02 \, \text{mol}
\]

\[
n_{\text{SO}_2} = \frac{V}{22,4} = \frac{7,437}{22,4} = 0,332 \, \text{mol}
\]

\[
n_{\text{HCl}} = C \cdot V = 0,5 \cdot 0,5 = 0,25 \, \text{mol}
\]

\[
n_{\text{CO}_2} = \frac{N}{N_A} = \frac{1,8066 \cdot 10^{22}}{6,022 \cdot 10^{23}} = 0,03 \, \text{mol}
\]

\[
n_{\text{H}_2\text{SO}_4} = \frac{m}{M} = \frac{4,9}{98} = 0,05 \, \text{mol}
\]

\[
m_{\text{NaCl}} = \frac{23,4}{100} \cdot 250 = 58,5 \, \text{gam}
\]
\[
n_{\text{NaCl}} = \frac{m_{\text{NaCl}}}{M_{\text{NaCl}}} = \frac{58,5}{58,5} = 1,0 \, \text{mol}
\]

\[
n_{\text{K}_2\text{SO}_4} = C \cdot V = 0,25 \cdot 2 = 0,5 \, \text{mol}
\]

\[
n_{\text{O}_2} = \frac{V}{22,4} = \frac{6,1975}{22,4} = 0,2765 \, \text{mol}
\]