Em tham khảo nha:
\(\begin{array}{l}
6)\\
a)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{{3,1}}{{31}} = 0,1\,mol\\
{n_{{O_2}}} = \dfrac{5}{{32}} = 0,15625\,mol\\
\dfrac{{{n_P}}}{4} < \dfrac{{{n_{{O_2}}}}}{5} \Rightarrow {O_2} \text{ dư }\\
{n_{{O_2}}} \text{ dư }= 0,15625 - 0,1 \times \dfrac{5}{4} = 0,03125\,mol\\
{m_{{O_2}}} \text{ dư } = 0,03125 \times 32 = 1g\\
b)\\
{n_{{P_2}{O_5}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{m_{{P_2}{O_5}}} = 0,05 \times 142 = 7,1g\\
7)\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
\text{ Để hỗn hợp nổ mạnh nhất } \Rightarrow {n_C}:{n_{{O_2}}} = 1:1\\
8)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{F{e_3}{O_4}}} = \dfrac{{2,32}}{{232}} = 0,01\,mol\\
{n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,03\,mol\\
{n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,02\,mol\\
{m_{Fe}} = 0,03 \times 56 = 1,68g\\
{m_{{O_2}}} = 0,02 \times 32 = 0,64g
\end{array}\)
