a,
$n_{Br_2}=n_{\text{anken}}=0,05(mol)$
$n_{hh}=\dfrac{3,36}{22,4}=0,15(mol)$
$\to n_{\text{ankan}}=0,15-0,05=0,1(mol)$
Đặt CTTQ ankan là $C_nH_{2n+2}$, anken là $C_nH_{2n}$
$\overline{M}_{hh}=21,66.2=43,32$
$\to 0,1(14n+2)+0,05.14n=43,32.0,15$
$\to n=3$
Vậy CTPT ankan là $C_3H_8$, anken là $C_3H_6$
b,
$3,36l$ hh có $0,1$ mol ankan, $0,05$ mol anken
$\to 4,48l$ hh có $\dfrac{2}{15}$ mol ankan, $\dfrac{1}{15}$ mol anken
$C_3H_8+5O_2\xrightarrow{{t^o}} 3CO_2+4H_2O$
$C_3H_6+\dfrac{9}{2}O_2\xrightarrow{{t^o}} 3CO_2+3H_2O$
$\to n_{O_2}=5n_{C_3H_8}+\dfrac{9}{2}n_{C_3H_6}=\dfrac{29}{30}(mol)$
$\to V_{kk}=5V_{O_2}=108,267l$
