Question

Giup em 1 bai nay vs mn

Answer 1

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

           0,05<--0,05

=> \(\%V_{C_2H_4}=\dfrac{0,05.22,4}{4,48}.100\%=25\%\)

\(\%V_{CH_4}=100\%-25\%=75\%\)