Đáp án:

Giải thích các bước giải:

Bài 4:

`n_{CO_2}=\frac{6,72}{22,4}=0,3(mol)`

`CO_2+Ba(OH)_2→BaCO_3↓+H_2O`

`0,3:0,3:0,3:0,3`

$a)$`C_{M_{Ba(OH)_2}}=\frac{0,3}{0,6}=0,5(M)`

$b)$`m_{BaCO_3}=0,3.197=59,1(g)`

Bài 5:

`n_{HCl}=0,3.1=0,3(mol)`

`n_{NaOH)=0,2.0,5=0,1(mol)`

`HCl+NaOH→NaCl+H_2O`

`\frac{0,3}1>\frac{0,1}1⇒`$HCl\ textit{dư},\ NaOH\ \textit{hết}$

`0,1:0,1:0,1:0,1`

$a)$`m_{HCl\ \textit{dư}}=36,5(0,3-0,1)=7,3(g)`

$b)$`m_{NaCl}=0,1.58,5-5,85(g)`

Bài 6:

`n_{MgO}=\frac{6}{40}=0,15(mol)`

`MgO+H_2SO_4→MgSO_4+H_2O`

`0,15:0,15:0,15:0,15`

$a)$`m_{H_2SO_4}=0,15.98=14,7(g)`

$b)$m_{ddH_2SO_4}=50.1,1=55(g)`

`C%_{ddH_2SO_4}=\frac{14,7}{55}.100=26,727%`

`C_{M_{ddH_2SO_4}}=\frac{0,15}{0,05}=3(M)`

Xin hay nhất!!!

4.

Ba(OH)2+CO2→BaCO3+H2O

nCO2=6,72/22,4=0,3mol

CM Ba(OH)2=0,3/0,6=0,5M

mBaCO3=0,3.197=59,1g

5.

NaOH+HCl→NaCl+H2O

nHCl=0,3mol

nNaOH=0,2.0,5=0,1mol

Vì pứ theo tỉ lệ 1:1

nHCl>nNaOH→HCl dư

nHCl dư=0,3-0,1=0,2mol

mHCl dư=0,2.36,5=7,3g

mNaCl=0,1.58,5=5,85g

6.

MgO+H2SO4→MgSO4+H2O

nMgO=6/40=0,15mol

mdd H2SO4=50.1,1=55g

mH2SO4 pứ=0,15.98=14,7g

mdd sau pứ=6+55=61g

mMgSO4=0,15.120=18g

C%MgSO4=18/61.100%=29,51%

CM MgSO4=0,15/0,05=3M