Đáp án:

\(\begin{array}{l}
b)\\
{V_{{O_2}}} = 3,36l\\
c)\\
{m_{A{l_2}{O_3}}} = 10,2g\\
d)\\
{m_{{P_2}{O_5}}} = 8,52g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
b)\\
{n_{KCl{O_3}}} = \dfrac{{12,25}}{{122,5}} = 0,1\,mol\\
{n_{{O_2}}} = 0,1 \times \dfrac{3}{2} = 0,15\,mol\\
{V_{{O_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
4Al + 3{O_2} \xrightarrow{t^0}2A{l_2}{O_3}\\
{n_{A{l_2}{O_3}}} = 0,15 \times \dfrac{2}{3} = 0,1\,mol\\
{m_{A{l_2}{O_3}}} = 0,1 \times 102 = 10,2g\\
d)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{{6,2}}{{31}} = 0,2\,mol\\
\dfrac{{{n_P}}}{4} > \dfrac{{{n_{{O_2}}}}}{5} \Rightarrow P \text{ dư }\\
{n_{{P_2}{O_5}}} = 0,15 \times \dfrac{2}{5} = 0,06\,mol\\
{m_{{P_2}{O_5}}} = 0,06 \times 142 = 8,52g
\end{array}\)

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