$m_{BaCl_{2}}$ `=` `(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`
