Đáp án:
CTHH của A có dạng $C_xH_y$
$C_xH_y+\bigg(x+\dfrac y4\bigg)\to xCO_2+\dfrac y2 H_2O$
$CO_2+Ca(OH)_2→CaCO_3(↓)+H_2O$
$n_{CaCO_3}=\dfrac{15}{100}=0,15(mol)$
$\to n_{CO_2}=n_{CaCO_3}=0,15(mol)$
$\to m_{CO_2}=0,15.44=6,6(g)$
$Δm_{binh}=12(g)$
$→m_{H_2O}=12-6,6=5,4(g)$
$\to n_{H_2O}=\dfrac{5,4}{18}=0,3(mol)$
$n_{H_2O}>n_{CO_2}→A$ là ankan
$→n_A=n_{H_2O}-n_{CO_2}=0,3-0,15=0,15(mol)$
$\to x=\dfrac{0,15}{0,15}=1$
$\to y=4$
$\to CH_4$