\(Butan\rightarrow Ankan+Anken\)
\(\rightarrow\) mX=m butan ban đầu
Mà \(M_X=11,6MH_2=11,6.2=23,2\)
\(\rightarrow\frac{n_X}{n_{Butan}}=\frac{M_{Butan}}{M_X}=\frac{58}{23,2}=2,5\)
Ta có: \(n_X=0,1\left(mol\right)\rightarrow n_{Butan_{bđ}}=\frac{0,1}{2,5}=0,04\left(mol\right)\)
\(n_{Anken_{tao.ra}}=n_X-n_{Butan}=0,1-0,04=0,06=nBr_2\)
\(\rightarrow m_{Br2}=0,06.160=9,6\left(g\right)\)