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$\rm 1.$ 

$\rm n_{CO_2}=$ $\rm\dfrac{143}{44}=3,25(mol)$ 

$-$ Bảo toàn nguyên tố C: 

$\rm n_{C}=n_{CH_4}=x(mol)$

$\rm n_C=4n_{C_4H_{10}}=4y(mol)$

$\rm\rightarrow x+4y=3,25(g)$

Mà $\rm m_{CH_4}+m_{C_4H_{10}}=47,5$

$\rm\rightarrow 16x+58y=47,5(2)$

$\rm (1),(2) \rightarrow x=0,25, y=0,75$

$-$ Bảo toàn nguyên tố H:

$\rm n_{H}=4n_{CH_4}=4.0,25=1(mol)$

$\rm n_{H}=10n_{C_4H_{10}}=10.0,75=7,5(mol)$

$\rm\rightarrow n_H=8,5(mol)$

$\rm\rightarrow n_{H_2}=4,25$ 

$\rm n_{H_2O}=n_{H_2}=4,25(mol)$

$\rm\rightarrow m=m_{H_2O}=4,25.18=76,5(g)$

$\rm b)$ $\rm \%V_{CH_4}=$ $\rm\dfrac{0,25}{0,25+0,75}.100\%=25\%$ 

$\rm\rightarrow \%V_{C_4H_{10}}=100\%-25\%=75\%$

$\rm c)$ KLM trung bình của hh khí gaz

$\rm M=$ $\rm\dfrac{47,5}{1}=47,5(g/mol)$ 

$\rm d_{HH/KK}=$ $\rm\dfrac{47,5}{29}\approx 1,64$ 

Bài 1:

`a)`

`n_{CO_2}={143}/{44}=3,25(mol)`

Đặt `n_{CH_4}=x(mol);n_{C_4H_{10}}=y(mol)`

Bảo toàn C: `x+4y=n_{CO_2}=3,25(1)`

Mà `16x+58y=47,5(2)`

`(1)(2)->x=0,25;y=0,75`

Bảo toàn H: `n_{H_2O}=2x+5y=4,25(mol)`

`->m=4,25.18=76,5(g)`

`b)`

`\%V_{CH_4}={0,25}/{0,25+0,75}.100\%=25\%`

`\%V_{C_4H_{10}}=100-25=75\%`

`c)\ d_{hh//kk}={M_{hh}}/{29}={{47,5}/{0,25+0,75}}/{29}\approx 1,64`