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)`
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