Đáp án:
$a)$
$d_{A/H_2}=4,5\to \overline{M_A}=4,5.2=9(g/mol)$
$n_{A}=\dfrac{4,48}{22,4}=0,2(mol)$
$H_2: 2$ 7
$9$
$CH_4:16$ 7
$\to n_{H_2}=n_{CH_4}=0,1(mol)$
$d_{B/H_2}=\dfrac{73}{6}\to \overline{M_B}=\dfrac{73}{6}.2=\dfrac{73}{3}(g/mol)$
$\to \dfrac{0,1.2+0,1.16+32x}{0,2+x}=\dfrac{73}{3}(g/mol)$
$\to x=0,4(mol)$
$\to V_{O_2}=0,4.22,4=8,96(l)$
$b)$
$2H_2+O_2\to 2H_2O$
$CH_4+2O_2\to CO_2+2H_2O$
$n_{H_2O}=0,2+0,1=0,3(mol)$
$\to m_{H_2O}=0,3.18=5,4(g)$