Đáp án:

$a\big)$

$400ml=0,4l\to n_{HCl}=2.0,4=0,8(mol)$

Bảo toàn H: $n_{H_2O}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,4(mol)$

Bảo toàn O: $n_O=n_{H_2O}=0,4(mol)$

$\to m_{X(P2)}=m_{oxit}-m_O=m_Y-m_O=14,2-0,4.16=7,8(g)$

$\to m_X=7,8.2=15,6(g)$

$b\big)$

$Mg+2HCl\to MgCl_2+H_2$

$2B+2bHCl\to 2BCl_b+bH_2$ 

Theo PT: $n_{Mg}+\dfrac{b.n_B}{2}=n_{H_2}=0,4$

$\to \dfrac{n_B}{2}+\dfrac{b.n_B}{2}=0,4$

$\to n_B.(b+1)=0,8$

$\to n_B=\dfrac{0,8}{b+1}(mol)$

$\to n_{Mg}=\dfrac{0,4}{b+1}(mol)$

$\to m_{Mg}=\dfrac{9,6}{b+1}(mol)$

$\to m_B=7,8-\dfrac{9,6}{b+1}=\dfrac{7,8b-1,8}{b+1}(g)$

$\to M_B=\dfrac{7,8b-1,8}{0,8}(g/mol)$

$\to \begin{cases} b=3\\ M_B=27\to B:Al \end{cases}$

Vậy $B$ là nhôm

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