a)
$C_2H_4 + Br_2 \to C_2H_4Br_2$
$n_{C_2H_4} = n_{Br_2} = \dfrac{32}{160} =0,2(mol)$
$\%V_{C_2H_4} = \dfrac{0,2.22,4}{6,72}.100\% = 66,67\%$
$\%V_{CH_4} = 100\% -66,67\% = 33,33\%$
b)
$n_{CH_4} = 0,1(mol)$
Bảo toàn C :
$n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = 0,5(mol)$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,5(mol)$
$m_{CaCO_3} = 0,5.100 = 50(gam)$