a,
$n_{C_2H_6}= a$
$n_{C_3H_6}= b$
=> $30a+ 42b= 8,22$ (1)
$n_{C_3H_6Br_2}= \frac{23,23}{202}= 0,115 mol= n_{C_3H_6}$ (bảo toàn C)
=> $b= 0,115$ (2)
(1)(2) => $a=0,113; b=0,115$
$\%m_{C_2H_6}= \frac{30.0,113.100}{8,22}= 41,24\%$
=> $\%m_{C_3H_6}= 58,76\%$
b,
Bảo toàn C:
$n_{CaCO_3 \downarrow}= n_{CO_2}= 2n_{C_2H_6}+ 3n_{C_3H_6}= 0,571mol$
=> $m_{\downarrow}= 0,571.100= 57,1g$