`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`
