Em tham khảo nha:
\(\begin{array}{l}
8)\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
{n_{{C_4}{H_{10}}}} = \dfrac{{13,05}}{{58}} = 0,225\,mol\\
{n_{{O_2}}} = \dfrac{{13}}{2} \times {n_{{C_4}{H_{10}}}} = 0,225 \times \dfrac{{13}}{2} = 1,4625\,mol\\
{V_{kk}} = 1,4625 \times 22,4 \times 5 = 163,8l\\
9)\\
a)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
{n_{A{l_2}{O_3}}} = \dfrac{{61,2}}{{102}} = 0,6mol\\
{n_{{O_2}}} = \dfrac{{0,6 \times 3}}{2} = 0,9\,mol\\
{V_{{O_2}}} = 0,9 \times 22,4 = 20,16l\\
b)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 0,9 \times \dfrac{2}{3} = 0,6\,mol\\
{m_{KCl{O_3}}} = 0,6 \times 122,5 = 73,5g
\end{array}\)
