a,

$BaCl_2+H_2SO_4\to BaSO_4+2HCl$

$n_{BaCl_2}=\dfrac{50.20,8\%}{208}=0,05(mol)$

$\Rightarrow n_{H_2SO_4}=n_{BaSO_4}=0,05(mol)$

$m=m_{dd H_2SO_4}=0,05.98:19,6\%=25g$

b,

$m_{dd\text{spu}}=25+50-0,05.233=63,35g$

$n_{HCl}=0,05.2=0,1(mol)$

$\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5.100}{63,35}=5,8%$

c,

$NaOH+HCl\to NaCl+H_2O$

$\Rightarrow n_{NaOH}=0,1(mol)$

$m_{dd NaOH}=0,1.40:5\%=80g$

$\Rightarrow V_{dd NaOH}=\dfrac{80}{1,2}=66,67ml$

 $m_{BaCl_{2}}$ `=` `(20,8.50)/100` `=10,4(g)`

`⇒` $n_{BaCl_{2}}$ `=` `(10,4)/208` `=0,05(mol)`

 $BaCl_{2}$ `+`  $H_{2}$$SO_{4}$  `→` $BaSO_{4}$`↓` `+` `2HCl`
`0,05(mol)`--→`0,05(mol)`-→`0,05(mol)`--→`0,01(mol)`

`a)`

$m_{H_{2}SO_{4}}$ `= 0,05.98=4,9(g)`

 Vậy `m=4,9(g)`

`b)`

$m_{HCl}$ `= 0,05.36,5=3,65(g)`

$m_{ddH_{2}SO_{4}}$ `=` `(4,9.100)/(19,6)``=25(g)`

`⇒` $m_{BaSO_{4}}$ `=233.0,05=11,65(g)`

`⇒`$m_{ddHCl}$ `=` `50+25-11,65=63,35(g)`   (định luật bảo toàn khối lượng)

`C%`$_{HCl}$ `=` `(3,65.100)/(63,35)``=5,8%`

`c)`

`NaOH + HCl` `→` `NaCl` + $H_{2}O$ 
0,1(mol)←0,1(mol)→`0,1(mol)→0,1(mol)

`⇒`$m_{ NaOH}$ `=` `40.0,1=4(g)`

`⇒`$m_{ ddNaOH}$ `=` `(4.100)/5``=80(g)`

`⇒` $V_{ ddNaOH}$ `=` `m_{ ddNaOH}/D``=` `80/(1,2)`` ≈``66,679(ml)`

`cuthilien`