Giải thích các bước giải:

\(\begin{array}{l}
7.\\
C + {O_2} \to C{O_2}\\
{n_C} = 0,4mol\\
{n_{{O_2}}} = 0,3mol\\
 \to {n_C} > {n_{{O_2}}} \to {n_C}dư\\
 \to {n_C}(pt) = {n_{{O_2}}} = 0,3mol\\
 \to {n_C}(dư) = 0,1mol \to {m_C}(dư) = 1,2g\\
 \to {n_{C{O_2}}} = {n_{{O_2}}} = 0,3mol \to {V_{C{O_2}}} = 6,72l\\
9.\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = 0,2mol\\
{n_{{O_2}}} = 0,3mol\\
 \to {n_P} < {n_{{O_2}}} \to {n_{{O_2}}}dư\\
 \to {n_{{P_2}{O_5}}} = \dfrac{1}{2}{n_P} = 0,1mol\\
 \to {m_{{P_2}{O_5}}} = 14,2g\\
10.\\
4K + {O_2} \to 2{K_2}O\\
{n_K} = 0,4mol\\
{n_{{O_2}}} = 0,2mol\\
 \to {n_K} < {n_{{O_2}}} \to {n_{{O_2}}}du\\
 \to {n_{{O_2}}}(pt) = \dfrac{1}{4}{n_K} = 0,1mol\\
a)\\
{n_{{K_2}O}} = \dfrac{1}{2}{n_K} = 0,2mol\\
 \to {m_{{K_2}O}} = 18,8g\\
b)\\
{O_2} + {H_2} \to {H_2}O\\
{n_{{O_2}}}(dư) = 0,1mol\\
 \to {n_{{H_2}}} = {n_{{O_2}}}(dư) = 0,1mol\\
 \to {m_{{H_2}}} = 0,2g
\end{array}\)