Đáp án:
$n_{H_2O}=\dfrac{5,4}{18}=0,3(mol)$
$\to n_H=0,3\times 2=0,6(mol)$
$\to m_H=0,6\times 1=0,6(g)$
$\to n_C=\dfrac{3-0,6}{12}=0,2(mol)$
A có dạng $C_xH_y$
$\to x:y=0,2:0,6=1:3$
$\to CTTQ: {(CH_3)}_n$
$d_{A/H_2}=15\to M_A=15.2=30(g/mol)$
$\to M_{{(CH_3)}_n}=30(g/mol)$
$\to 15n=30$
$\to n=2$
$\to C_2H_6$