$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

$PTHH$ $:$

$4Na$ $+$ $O_2$ ${\text{ }}\xrightarrow{{{t^o}}}$ $2Na_2 O$ $(1)$

$2Mg$ $+$ $O_2$ ${\text{ }}\xrightarrow{{{t^o}}}$ $2MgO$ $(2)$

$4Al$ $+$ $3O_2$ ${\text{ }}\xrightarrow{{{t^o}}}$ $2Al_2 O_3$ $(3)$

$2Na$ $+$ $2HCl$ $→$ $2NaCl$ $+$ $H_2↑$ $(4)$

$Mg$ $+$ $2HCl$ $→$ $MgCl_2$ $+$ $H_2↑$ $(5)$

$2Al$ $+$ $6HCl$ $→$ $2AlCl_3$ $+$ $3H_2↑$ $(6$

$Theo$ $pt$ $(1,2,3)$ $:$

$Áp$ $dụng$ $định$ $luật$ $bảo$ $toàn$ $nguyên$ $tố$ $:$

$m_{O}$ $=$ $m_{oxit}$ $-$ $m_{kim}$ $_{loại}$ $=$ $17$ $-$ $10,2$ $=$ $6,8$ $(g)$

$⇒$ $n_{O}$ $=$ $\frac{6,8}{16}$ $=$ $0,425$ $(mol)$

$Theo$ $pt$ $(1,2,3,4,5,6)$ $:$ $n_{H_2}$ $=$ $n_{O}$ $=$ $0,425$ $(mol)$

$⇒$ $V_{H_2}$ $=$ $0,425$ $×$ $22,4$ $=$ $9,52$ $(lít)$

$Theo$ $pt$ $(1,2,3,4,5,6)$ $:$ $n_{Cl}$ $=$ $2×n_{O}$ $=$ $0,85$ $(mol)$

$⇒$ $m_{muối}$ $=$ $m$ $=$ $m_{kim}$ $_{loại}$ $+$ $m_{Cl}$ $=$ $10,2$ $+$ $0,85×35,5$ $=$ $40,375$ $(g)$

Đáp án + Giải thích các bước giải:

Bảo toàn KL:

$m_{O_2}=17-10,2=6,8(g)$

$\to n_{O_2}=\frac{6,8}{32}=0,2125(mol)$

$\to n_{H_2}=n_{H_2O}=2n_{O_2}=0,425(mol)$

$\to V=V_{H_2}=0,425.22,4=9,52(l)$

Ta có $n_{Cl^-}=n_{HCl}=2n_{H_2}=0,85(mol)$

$\to m=m_{muối}=m_{KL}+m_{Cl^-}=0,85.35,5+10,2=40,375(g)$