a) $C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O$
b) Theo PTHH : $V_{O_2} = \dfrac{13}{2}V_{C_4H_{10}} = 21,84(lít)$
$n_{C_4H_{10}} = 0,15(mol) \Rightarrow n_{H_2O} = 5n_{C_4H_{10}} = 0,75(mol)$
$m = 0,75.18 = 13,5(gam)$
c) $2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$n_{KMnO_4\ pư} = 2n_{O_2} = 1,95(mol)$
$n_{KMnO_4\ đã\ dùng} = 1,95 : 80\% = 2,4375(mol)$
$m_{KMnO_4} = 2,4375.158 = 385,125(gam)$
\(\begin{array}{l}
a)\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
b)\\
{n_{{C_4}{H_{10}}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{{O_2}}} = 0,15 \times \dfrac{{13}}{2} = 0,975\,mol\\
{V_{{O_2}}} = 0,975 \times 22,4 = 21,84l\\
{m_{{O_2}}} = 0,975 \times 32 = 31,2g\\
c)\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 1,95\,mol\\
{m_{KMn{O_4}}} = \dfrac{{1,95 \times 158}}{{80\% }} = 385,125g
\end{array}\)