a,
Propilen bị dd brom hấp thụ.
$\Rightarrow m_{\text{tăng}}=m_{C_3H_6}=6,3g$
$n_{C_3H_6}=\dfrac{6,3}{42}=0,15(mol)$
$m_{CH_4}=10,3-6,3=4g$
$\Rightarrow n_{CH_4}=\dfrac{4}{16}=0,25(mol)$
$\%V_{CH_4}=\dfrac{0,25.100}{0,25+0,15}=62,5\%$
$\to \%V_{C_3H_6}=37,5\%$
b,
Trong $Z$ có $0,15$ mol propilen.
$\Rightarrow \dfrac{1}{4}$ $Z$ có $\dfrac{0,15}{4}=0,0375(mol)$
$3C_3H_6+2KMnO_4+4H_2O\to 3C_3H_6(OH)_2+2MnO_2+2KOH$
$\Rightarrow n_{KMnO_4}=\dfrac{2}{3}n_{C_3H_6}=0,025(mol)$
$\to V_{KMnO_4}=\dfrac{0,025}{2}=0,0125l=12,5ml$
