`#Ly`

Gọi kim loại hóa trị $II$ là $A$

Đổi $250$ $ml$ $=$ $0,25$ $l$

Đổi $60$ $ml$ $=$ $0,06$ $l$

$n_{H_2SO_4}$ $=$ `C_M.V=0,3.0,25=0,075` $(mol)$

$n_{NaOH}$ $=$ `C_M.V=0,5.0,06=0,03` $(mol)$

PTHH:

$(1):$ $A$ $+$ $H_2SO_4$ `->` $ASO_4$ $+$ $H_2↑$

$(2):$ $2NaOH$ $+$ $H_2SO_4$ `->` $Na_2SO_4$ $+$ $2H_2O$

Theo PT $(2)$, $n_{H_2SO_4}$ $=$ $\dfrac{1}{2}n_{NaOH}$ $=$ $0,015$ $(mol)$

$⇒$ $n_{H_2SO_4(1)}$ $=$ $0,075-0,015=0,06$ $(mol)$

Theo PT $(1)$, $n_{H_2SO_4}$ $=$ $n_{A}$ $=$ $0,06$ $(mol)$

Mà theo đề bài, $n_{A}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{1,44}{A}$ $(mol)$

$⇒$ $\dfrac{1,44}{A}$ $=$ $0,06$

$⇒$ $A=24$

Vậy $A$ là kim loại $Mg$

_____ _____ _____ _____ _____ _____ _____ _____ _____

Đổi $300$ $ml$ $=$ $0,3$ $l$

Đổi $200$ $ml$ $=$ $0,2$ $l$

Đổi $500$ $ml$ $=$ $0,5$ $l$

$n_{HCl(bđ)}$ $=$ `C_M.V=0,5.0,3=0,15` $(mol)$

$n_{HCl(spư)}$ $=$ `C_M.V=0,5.0,02=0,01` $(mol)$

$n_{HCl(pư)}$ $=$ $n_{HCl(bđ)}$ $-$ $n_{HCl(spư)}$ $=$ $0,15-0,01=0,14$ $(mol)$

PTHH: $2HCl$ $+$ $Ba(OH)_2$ `->` $BaCl_2$ $+$ $2H_2O$

Theo PT, $n_{HCl(pư)}$ $=$ $\dfrac{1}{2}n_{Ba(OH)_2}$ $=$ $0,07$ $(mol)$

`C_MBa(OH)_2` $=$ $\dfrac{n}{V}$ $=$ $\dfrac{0,07}{0,2}$ $=$ $0,35$ $(M)$

\(\begin{array}{l}
a)\\
250ml=0,25l;60ml=0,06l\\
n_{H_2SO_4}=0,25.0,3=0,075(mol)\\
n_{NaOH}=0,06.0,5=0,03(mol)\\
2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\
Theo\,PT:\,n_{H_2SO_4(dư)}=\frac{1}{2}n_{NaOH}=0,015(mol)\\
\to n_{H_2SO_4(pứ)}=0,075-0,015=0,06(mol)\\
Dat\,kim\,loai\,la\,R\\
R+H_2SO_4\to RSO_4+H_2\\
Theo\,PT:\,n_R=n_{H_2SO_4}=0,06(mol)\\
\to M_R=\frac{1,44}{0,06}=24(g/mol)\\
\to R\,la\,magie\,(Mg)\\
b)\\
300ml=0,3l;200ml=0,2l;500ml=0,5l\\
n_{HCl(bđ)}=0,3.0,5=0,15(mol)\\
n_{HCl(dư)}=0,5.0,02=0,01(mol)\\
\to n_{HCl(pứ)}=0,15-0,01=0,14(mol)\\
Ba(OH)_2+2HCl\to BaCl_2+2H_2O\\
Theo\,PT:\,n_{Ba(OH)_2}=\frac{1}{2}n_{HCl}=0,07(mol)\\
\to a=C_{M\,Ba(OH)_2}=\frac{0,07}{0,2}=0,35M
\end{array}\)