$n_{C2H4Br2}$=$\frac{1,7}{188}$≈0,009 (mol)
a, CH4 + Br2 ∦ Không phản ứng
C2H4 + Br2 -> C2H4Br2
0,009 0,009 0,009
b, $m_{Br2}$=0,009.160=1,44 (g)
c, $V_{C2H4}$=0,009.22,4=0,2016 (l)
⇒%$V_{C2H4}$=$\frac{0,2016.100}{3}$=6,72%
⇒%$V_{CH4}$=100%-6,72%=93,28%