Cracking butan thu được hổn hợp T gồm 7 chất: CH4, C3H6, C2H4, C2H6, C4H8, H2, C4H6.

Vì là phản ứng cracking nên:
\(\\ n_{CH_{4}} = n_{C_{3}H_{6}} ; \ n_{C_{2}H_{4}} = n_{C_{2}H_{6}} ; \ n_{C_{4}H_{8}}= n_{H_{2}}; \ n_{C_{4}H_{6}}.2 = n_{H_{2}} \\ \Rightarrow n_{CO_{2}}= 4n_{C_{4}H_{10}}= 4(n_{C_{3}H_{6}} + n_{C_{2}H_{4}}+ n_{C_{4}H_{6}} + n_{C_{4}H_{8}} ) \\ \Rightarrow n_{C_{3}H_{6}} + n_{C_{2}H_{4}} + n_{C_{4}H_{6}} + n_{C_{4}H_{8} }= 0,1 \ mol \\ n_{Br_{2}} = n_{C_{3}H_{6}} + n_{C_{2}H_{4}} + n_{C_{4}H_{6}} + n_{C_{4}H_{8}} = 0,12 \ mol \\ \Rightarrow n_{C_{4}H_{6}} = 0,02 \ mol\)
Bảo toàn khối lượng:
\(​ \\ m_{T} = m_{C_{4}H_{10} \ bd} = 0,1.58 = 5,8 \ g \\ \Rightarrow \%m_{C_{4}H_{6} \ (T)} = 18,62 \ \%\)\(​ \\ m_{T} = m_{C_{4}H_{10} \ bd} = 0,1.58 = 5,8 \ g \\ \Rightarrow \%m_{C_{4}H_{6} \ (T)} = 18,62 \ \%\)\(\\ m_{T} = m_{C_{4}H_{10} \ bd} = 0,1.58 = 5,8 \ g \\ \Rightarrow \%m_{C_{4}H_{6} \ (T)} = 18,62 \ \%\)