Đáp án:

\(\begin{array}{l}
b)\\
{m_P} = 0,031g\\
c)\\
{m_{{P_2}{O_5}}} = 11,36g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
b)\\
{n_P} = \dfrac{m}{M} = \dfrac{5}{{31}} \approx 0,161mol\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
\dfrac{{0,2}}{5} < \dfrac{{0,161}}{4} \Rightarrow Photpho\text{ dư}\\
{n_{{P_d}}} = {n_P} - \dfrac{4}{5}{n_{{O_2}}} = 0,001mol\\
{m_P} = n \times M = 0,001 \times 31 = 0,031g\\
c)\\
{n_{{P_2}{O_5}}} = \dfrac{{2{n_{{O_2}}}}}{5} = 0,08mol\\
{m_{{P_2}{O_5}}} = n \times M = 0,08 \times 142 = 11,36g
\end{array}\)