Đáp án:
Giải thích các bước giải:
`3.`
`n_{H_2SO_4}=0,4.0,5=0,2(mol)`
`n_{HCl}=0,4.1=0,4(mol)`
`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`
`NaOH+HCl->NaCl+H_2O`
`∑n_{NaOH}=2.0,2+0,4=0,8(mol)`
`V_{ddNaOH}=(0,8)/(0,75)=1,067(l)=1066,667(ml)`
`4.`
`a.`
`n_{MgCl_2}=0,1.2=0,2(mol)`
`n_{Ba(OH)_2}=0,15.1,5=0,225(mol)`
`MgCl_2+Ba(OH)_2->Mg(OH)_2↓+BaCl_2`
`(0,2)/1<(0,225)/1=>Ba(OH)_2` dư
`n_{Mg(OH)_2}=n_{MgCl_2}=0,2(mol)`
$Mg(OH)_2\xrightarrow{t^o}MgO+H_2O$
`n_{MgO}=n_{Mg(OH)_2}=0,2(mol)`
`m_{MgO}=0,2.40=8(g)`
`b.`
`n_{Ba(OH)_2pư}=n_{MgCl_2}=0,2(mol)`
`n_{Ba(OH)_2dư}=0,225-0,2=0,025(mol)`
`n_{BaCl_2}=n_{MgCl_2}=0,2(mol)`
`V_A=0,1+0,15=0,25(l)`
`C_{M_{Ba(OH)_2}}=(0,025)/(0,25)=0,1(M)`
`C_{M_{BaCl_2}}=(0,2)/(0,25)=0,8(M)`
`m_A=250.1,12=280(g)`
`C%_{Ba(OH)_2}=(0,025.171}/280 .100=1,527%`
`C%_{BaCl_2}=(0,2.208)/280 .100=14,857%`
