Đáp án + Giải thích các bước giải:

$a\bigg)C_2H_2+2Br_2\to C_2H_2Br_4$

$b\bigg)$

Theo PT: $n_{C_2H_2}=\frac{1}{2}n_{Br_2}=\frac{1}{2}.\frac{32}{160}=0,1(mol)$

$\to \%V_{C_2H_2}=\frac{0,1.22,4}{6,72}.100\%=33,33\%$

$\to \%V_{CH_4}=100-33,33=66,67\%$

$c\bigg)$

$n_{CH_4}=\frac{6,72-0,1.22,4}{2,24}=0,2(mol)$

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

$2C_2H_2+5O_2\xrightarrow{t^o}4CO_2+2H_2O$

Theo PT: $\sum n_{O_2}=2n_{CH_4}+\frac{5}{2}n_{C_2H_2}=0,65(mol)$

$\to V_{O_2}=0,65.22,4=14,56(l)$

`1)`

`C_2H_2+ 2Br_2 → C_2H_2Br_4`

`2)`

`n_{Br_2} = {32}/{160} = 0,2(mol)` 

`n_{C_2H_2} = 1/2n_{Br_2} = 0,1(mol)` 

`%V_{C_2H_2}= {0,1 .22,4}/{6,72} . 100= 33,33%`

`%V_{CH_4} = 100-33,33= 66,67%`

`3)`

`CH_4 + 2O_2 → CO_2+ 2H_2O`

`2C_2H_2 + 5O_2→ 4CO_2 + 2H_2O` 

`V_{O_2} = 2V_{CH_4} + 5/2V_{C_2H_2}`

`= (6,72 - 0,1 .22,4 ).  2 + 5/2 . 0,1 .22,4`

`= 14,56(l)`