Giả sử $m_A=m_B=6,962g$
$\overline{M}_A=11,8.2=23,6$
$\Rightarrow n_A=\dfrac{6,962}{23,6}=0,295(mol)$
$\overline{M}_B=14,75.2=29,5$
$\Rightarrow n_B=\dfrac{6,962}{29,5}=0,236(mol)$
$\dfrac{n_{C_4H_8}}{n_{H_2}}=\dfrac{23,6-2}{56-23,6}=\dfrac{2}{3}$
$\%n_{C_4H_8}=\dfrac{2.100}{2+3}=40\%$
$\Rightarrow n_{C_4H_8}=0,295.40\%=0,118(mol)$
$n_{H_2}=0,295-0,118=0,177(mol)$
Ta có: $n_{\text{giảm}}=0,295-0,236=0,059(mol)=n_{H_2\text{pứ}}$
$\Rightarrow n_{C_4H_8\text{pứ}}=0,059(mol)$
Theo lí thuyết, but-1-en hết.
$\to H=\dfrac{0,059.100}{0,118}=50\%$
