$BT14$ $:$ 

$PTHH$ $:$

$AO$ $+$ $H_2 O$ $→$ $A(OH)_2$

$m_{dd}$ $=$ $500$ $+$ $11,2$ $=$ $511,2$ $(g)$

$m_{A(OH)_2}$ $=$ $511,2$ $×$ $2,895\text{%}$ $=$ $≈14,8$ $(g)$

$n_{A(OH)_2}$ $=$ $\frac{14,8}{A+17×2}$ $(mol)$

$n_{AO}$ $=$ $\frac{11,2}{A+16}$ $(mol)$

$Theo$ $pt$ $:$ $n_{AO}$ $=$ $n_{A(OH)_2}$ 

$⇒$ $\frac{11,2}{A+16}$ $=$ $\frac{14,8}{A+17×2}$

$⇒$ $A$ $=$ $40$ $(g/mol)$

$⇒$ $A$ $là$ $Canxi$ $(Ca)$

$BT15$ $:$ 

$PTHH$ $:$

$A$ $+$ $2H_2 O$ $→$ $A(OH)_2$ $+$ $H_2↑$

$n_{A}$ $=$ $\frac{13,7}{A}$ $(mol)$

$Theo$ $pt$ $:$ $n_{H_2}$ $=$ $n_{A}$ $=$ $\frac{13,7}{A}$ $(mol)$

$n_{H_2}$ $=$ $\frac{13,7×2}{A}$ $=$ $\frac{27,4}{A}$ $(g)$

$m_{dd}$ $=$ $350$ $+$ $13,7$ $-$ $\frac{27,4}{A}$ $=$ $363,7$ $-$ $\frac{27,4}{A}$ $(g)$

$Theo$ $pt$ $:$ $n_{A(OH)_2}$ $=$ $n_{A}$ $=$ $\frac{13,7}{A}$ $(mol)$

$m_{A(OH)_2}$ $=$ $\frac{13,7×(A+17×2)}{A}$ $=$ $\frac{13,7A+468,8)}{A}$ $(g)$

$⇒$ $\frac{\frac{13,7A+468,8)}{A}}{363,7-\frac{27,4}{A}}$ $=$ $4,7043\text{%}$

$⇒$ $A$ $≈$ $137$ $(g/mol)$

$⇒$ $A$ $là$ $Bari$ $(Ba)$

$BT16$ $:$ 

$PTHH$ $:$

$2X$ $+$ $2nHCl$ $→$ $2XCl_n$ $+$ $nH_2↑$

$n_{HCl}$ $=$ $\frac{300×7,3\text{%}}{36,5}$ $=$ $0,6$ $(mol)$

$Theo$ $pt$ $:$ $n_X$ $=$ $\frac{1}{n}×n_{HCl}$ $=$ $\frac{0,6}{n}$ $(mol)$

$⇒$ $M_X$ $=$ $\frac{5,4}{\frac{0,6}{n}}$ $=$ $9n$

$+$ $n=1$ $⇒$ $X$ $=$ $9$ $(loại)$

$+$ $n=2$ $⇒$ $X$ $=$ $18$ $(loại)$

$+$ $n=3$ $⇒$ $X$ $=$ $27$ $⇒$ $X$ $là$ $nhôm$ $(Al)$

$Theo$ $pt$ $:$ $n_{H_2}$ $=$ $\frac{1}{2}×n_{HCl}$ $=$ $0,3$ $(mol)$

$⇒$ $m_{\text{dd sau pứ}}$ $=$ $5,4$ $+$ $300$ $-$ $0,3$ $×$ $2$ $=$ $304,8$ $(g)$

$Theo$ $pt$ $:$ $n_{AlCl_3}$ $=$ $\frac{1}{3}×n_{HCl}$ $=$ $0,2$ $(mol)$

$⇒$ $n_{AlCl_3}$ $=$ $0,2$ $×$ $133,5$ $=$ $26,7$ $(g)$

$⇒$ $C\text{%}_{\text{dd sau pứ}}$ $=$ $\frac{26,7}{304,8}$ $×$ $100\text{%}$ $=$ $8,76\text{%}$

$Chúc$ $bn$ $hk$ $tốt$ $!!$ $:))))$ 

Đáp án $+$ Giải thích các bước giải:

Bài 14:

$n_{AO}=\dfrac{11,2}{A+16}(mol)$

$AO+H_2O→A(OH)_2$

Theo PT: $n_{A(OH)_2}=n_{AO}=\dfrac{11,2}{A+16}(mol)$

$C\%_{dd\ sau}=\dfrac{\dfrac{11,2}{A+16}×(A+34)}{11,2+500}×100\%=2,895\%$

$→A=40\ (Ca)$

Bài 15:

$n_{AO}=\dfrac{13,7}{A}(mol)$

$A+2H_2O→A(OH)_2+H_2$

Theo PT: $n_{A(OH)_2}=n_{AO}=\dfrac{13,7}{A}(mol)$

$n_{H_2}=\dfrac{13,7}{A}(mol)$

$C\%_{dd\ sau}=\dfrac{\dfrac{13,7}{A}×(A+34)}{13,7+350-2×\dfrac{13,7}{A}}×100\%=4,7043\%$

$→A=137\ (Ba)$

Bài 16:

Hóa trị $X:\ n$

$n_{HCl}=\dfrac{\dfrac{300×7,3}{100}}{36,5}=0,6(mol)$

$2X+2nHCl→2XCl_n+nH_2$

Theo PT: $n_X=\dfrac{2×n_{HCl}}{2n}=\dfrac{0,6}{n}(mol)$

$m_X=\dfrac{0,6}{n}×X=5,4(g)$

$⇒X=9n$

Với $n=3⇒X=27\ (Al)$

Theo PT: $n_{AlCl_3}=n_{Al}=\dfrac{0,6}{3}=0,2(mol)$

$n_{H_2}=\dfrac{n_{HCl}}{2}=0,3(mol)$

$m_{dd\ sau}=5,4+300-0,3×2=304,8(g)$

$→C\%_{dd\ sau}=\dfrac{0,2×133,5}{304,8}×100\%=8,76\%$