Giải thích các bước giải:

`a.`

`-` Vì `C_4H_{10}` chiếm `80%`

`→m_{C_4H_{10}}=m_{hh}.80%=20.80%=16(g)`

`→m_{CH_4}=m_{hh}-m_{C_4H_{10}}=20-16=4(g)`

`b.`

`-n_{C_4H_{10}}=\frac{m_{C_4H_{10}}}{M_{C_4H_{10}}}=\frac{16}{58}=\frac{8}{29}(mol)`

`-n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{4}{16}=0,25(mol)`

Phương trình hóa học :

$2C_4H_{10}+13O_2\xrightarrow{t^o}8CO_2↑+10H_2O$

`\frac{8}{29}`    `→`     `\frac{52}{29}`             `\frac{32}{29}`        `\frac{40}{29}`              `(mol)`   

 $CH_4+2O_2\xrightarrow{t^o}CO_2↑+2H_2O$

`0,25` `→`   `0,5`       `0,25`      `0,5`                `(mol)`

`→∑n_{O_2}=\frac{52}{29}+0,5=\frac{133}{58}(mol)`

`→m_{O_2}=n_{O_2}.M_{O_2}=\frac{133}{58}.32=\frac{2128}{29}≈73,38(g)`

`c.`

`-∑n_{CO_2}=\frac{32}{29}+0,25=\frac{157}{116}(mol)`

`→m_{CO_2}=n_{CO_2}.M_{CO_2}=\frac{157}{116}.44=\frac{1727}{29}≈59,55(g)`

`-∑n_{H_2O}=\frac{40}{29}+0,5=\frac{109}{58}(mol)`

`→m_{H_2O}=n_{H_2O}.M_{H_2O}=\frac{109}{58}.18=\frac{981}{29}≈33,83(g)`

$\boxed{\text{LOVE TEAM}}$

$\text{a) Ta có:}$

$m_{CH_4}=20.\dfrac{20}{100}=4\text{(g)}$

$m_{C_4H_{10}}=20.\dfrac{80}{100}=16\text{(g)}$

$\text{Phương trình:}$

$CH_4+ 2O_2\xrightarrow{t^0} CO_2 + 2H_2O$

$2C_4H_{10} + 13O_2 \xrightarrow{t^0} 8CO_2 + 10H_2O$

$\to n_{C_4H_{10}}=\dfrac{16}{58}=\dfrac{8}{29}\text{(mol)}$

$\to n_{CH_4}=\dfrac{4}{16}=0,25\text{(mol)}$

$\text{b)}$

$n_{O_2\text{pứ}}=\dfrac{8}{29}.\dfrac{13}{2}+ 0,25.\dfrac{6}{3}=\dfrac{133}{58}\text{(mol)}$

$\to m_{O_2\text{pứ}}=\dfrac{133}{58}.32=73,37\text{(g)}$

$\text{c)}$

$n_{CO_2}=\dfrac{8}{29}.\dfrac{8}{2}+0,25=\dfrac{157}{116}\text{(mol)}$

$\to V_{CO_2\text{pứ}}=\dfrac{157}{116}.22,4=30,32\text{(l)}$

$n_{H_2O}=\dfrac{8}{29}.5+ 0,25.2=\dfrac{109}{58}\text{(mol)}$

$\to m_{H_2O}=\dfrac{109}{58}.18=33,83\text{(g)}$