`n_{Br_2}=\frac{32}{160}=0,2(mol)`
`n_{\text{hh khí}}=\frac{6,2}{22,4}=\frac{31}{112}(mol)`
`a)` Phương trình:
`C_2H_2+2Br_2\to C_2H_2Br_4`
`=> n_{C_2H_2}=0,5n_{Br_2}=0,1(mol)`
`=> n_{CH_4}=\frac{31}{112}-0,1=\frac{99}{560}(mol)`
`b)` $\%V_{CH_4}=\dfrac{\dfrac{99}{560}.100\%}{\dfrac{31}{112}}\approx 63,87\%$
`%V_{C_2H_2}=\frac{0,1.100%}{\frac{31}{112}}\approx 36,13%`