Đáp án:
\(\begin{array}{l}
a)\\
{m_{{P_2}{O_5}}} = 5,68g\\
b)\\
{m_{{O_2}d}} = 0,8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
{n_P} = \dfrac{{2,48}}{{31}} = 0,08mol\\
{n_{{P_2}{O_5}}} = \dfrac{{{n_P}}}{2} = 0,04mol\\
{m_{{P_2}{O_5}}} = 0,04 \times 142 = 5,68g\\
b)\\
{n_{{O_2}}} = \dfrac{4}{{32}} = 0,125mol\\
4P + 5{O_2} \to 2{P_2}{O_5}\\
\dfrac{{0,08}}{4} < \dfrac{{0,125}}{5} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = {n_{{O_2}}} - \frac{5}{4}{n_P} = 0,025mol\\
{m_{{O_2}d}} = 0,025 \times 32 = 0,8g
\end{array}\)