Giả sử có 1 mol C4H10 tham gai phản ứng
\(M_X=9,0625.4=36,25\left(g/mol\right)\)
Theo ĐLBTKL: \(m_X=m_{C_4H_{10}}=58\left(g\right)\)
=> \(n_X=\dfrac{58}{36,25}=1,6\left(mol\right)\)
PTHH:
\(C_4H_{10}\xrightarrow[]{crackinh}CH_4+C_3H_6\)
\(C_4H_{10}\xrightarrow[]{crackinh}C_2H_6+C_2H_4\)
Ta có: \(n_{C_4H_{10}}=\Delta n_{\uparrow}=1,6-1=0,6\left(mol\right)\)
=> \(H=\dfrac{0,6}{1}.100\%=60\%\)