a) \(C2H4+Br2-->C2H4Br2\)
\(n_{Br}=0,3.1=0,3\left(mol\right)\)
\(n_{C2H4}=n_{Br2}=0,3\left(mol\right)\)
\(V_{C2H4}=0,3.22,4=6,72\left(l\right)\)
\(V_{CH4}=9,6-6,72=2,88\left(l\right)\)
b)\(\%V_{C2H4}=\frac{6,72}{9,6}.100\%=70\%\)
\(\%V_{CH4}=100-70=30\%\)
a) C2H4+Br2−−>C2H4Br2
0,3-----0,3 mol
nBr=0,3.1=0,3(mol)
=>VC2H4=0,3.22,4=6,72(l)
=>VCH4=9,6−6,72=2,88(l)
b)
-->%VC2H4=6,729,6.100%=70%
-->%VCH4=100−70=30%