Đáp án:

 40,8g

Giải thích các bước giải:

 \(\begin{array}{l}
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
hh:C{H_4}(a\,mol),{C_4}{H_{10}}(b\,mol)\\
{n_{C{O_2}}} = \dfrac{{33}}{{44}} = 0,75\,mol\\
\left\{ \begin{array}{l}
a + 4b = 0,75\\
16a + 58b = 11,1
\end{array} \right.\\
 \Rightarrow a = b = 0,15\,mol\\
\% {m_{C{H_4}}} = \dfrac{{0,15 \times 16}}{{11,1}} \times 100\%  = 21,62\% \\
\% {m_{{C_4}{H_{10}}}} = 100 - 21,62 = 78,38\% \\
{n_{{O_2}}} = 0,15 \times 2 + 0,15 \times 6,5 = 1,275\,mol\\
{m_{{O_2}}} = 1,275 \times 32 = 40,8g
\end{array}\)

`n_{CO_2}=\frac{33}{44}=0,75(mol)`

Cho ``$\begin{cases}CH_4: x(mol)\\ C_4H_{10 }: y(mol)\\\end{cases}$``

Ta có:

`m_{CH_4}+m_{C_4H_{10}}=11,1g`

`=> 16x+58y=11,1g(1)`

Phương trình:

`CH_4+2O_2\overset{t^o}{\to}CO_2+2H_2O`

`C_4H_10+\frac{13}{2}O_2\overset{t^o}{\to }4CO_2+5H_2O`

Theo phương trình, ta nhận thấy:

`n_{CO_2}=x+4y=0,75(mol)(2)`

Từ `(1), (2)` suy ra $\begin{cases} x=0,15(mol)\\y=0,15(mol)\\\end{cases}$

`=> %m_{CH_4}=\frac{16.0,15.100%}{11,1}\approx 21,62%`

`=> %m_{C_4H_10}=\frac{58.0,15.100%}{11,1}\approx 78,38%`

`n_{O_2}=2x+6,5y=2.0,15+6,5.0,15=1,275(mol)`

`=> m_{O_2}=1,275.32=40,8g`