1. $Δ_m=m_{O_2}=3,84(g)$
$\to n_O=\dfrac{3,84}{16}=0,24(mol)$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$
Khi cho X vào HCl
2H+ + O + 2e → H2O
2H+ + 2e → H2
$→∑n_{Cl-}=0,78(mol)$
$\to m=42,69-0,78.35.5=15(g)$
$\to B$
2. $n_O=\dfrac{20.25,6\%}{16}=0,32(mol)$
$→n_{H_2O}=0,32(mol)$
Bảo toàn H
$→n_{HCl}=2.n_{H_2O}+2.n_{H_2}=2.0,32+2.0,1=0,84(mol)$
$→m_{KL}=20-m_O=20-5,12= 14,88(g)$
$\to m=14,88+0,84.35,5=44,7$
$\to D$