$n_{HCl} = 0,4(mol)$
Ba(OH)₂ +  2HCl → BaCl₂ + 2H₂O

0,2................0,4.........0,2.........................(mol)

$x = \dfrac{0,4}{0,2} = 2M$

$V_{dd} = 0,4 + 0,2 = 0,6(mol)$
$C_{M_{BaCl_2}} = \dfrac{0,2}{0,6} = 0,33M$

a, PTHH: Fe2O3 + 6 HCl => 2FeCl3 + 3H2O.

Ta có: n_Fe2O3 =  0.1 mol.

Theo tỉ lệ pt ta có: n _HCl = 6n_Fe2O3  = 0.6 mol

=> a= 0.6/ 0.4 = 1.5M

b, dd sau pư là FeCl3: n _FeCl3 = 2n_Fe2O3 = 0.2 mol => C_M=0.2/ 0.4 = 0.5 mol.

\(n_{H2SO4}=0,25\left(mol\right)\)

\(PTHH:BaCl2+H2SO4\rightarrow BaSO4+2HCL\)

\(\Rightarrow\frac{V}{22,4}.1=0,25\Rightarrow V=5,6l\)

\(\Rightarrow n_{BaSO4}=0,25mol\Rightarrow m_{BaSO4}=58,25\)

\(\Rightarrow V_{BaSO4}=0,25.22,4=5,6l\)

\(\Rightarrow C_M=0,04M\)

a) BaCl2 + H2SO4--->BaSO4 +2HCl

Ta có

n\(_{H2SO4}=0,5.\)0,5=0,25(mol)

Theo pthh

nn\(_{BaCl2}=n_{H2SO4}=0,25\left(mol\right)\)

V=\(\frac{0,25}{1}=0,25\left(l\right)\)

b) Theo pthh

n\(_{BaSO4}=n_{H2SO4}=0,25\left(mol\right)\)

m\(_{BaSO4}=0,25.233=58,25\left(g\right)\)

Theo pthh

n\(_{HCl}=2n_{H2SO4}=0,5\left(mol\right)\)

C\(_{M\left(HCl\right)}=\frac{0,5}{0,5+0,25}=0,67\left(M\right)\)

Chúc bạn học tốt

Bài 1

\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bài 5

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

nBaO=30,6/153=0,2(mol)

a) PTHH: BaO + 2 HNO3 -> Ba(NO3)2 + H2O

b) nHNO3=0,2.2=0,4(mol) => mHNO3= 63.0,4=25,2(g)

=>C%ddHNO3=(25,2/150).100=16,8%

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,1-->0,2

=> \(C_{M\left(dd.HCl\right)}=\dfrac{0,2}{0,2}=1M\)

`n_(Zn) = (6,5)/65 = 0,1 (mol)`

`PTHH: ZN + 2HCl --> ZnCl_2 + H_2`

`0,1 --> 0,2`

`=> C_(M(dd.HCl) = (0,2)/(0,2) = 1M`

\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

Na₂CO₃  +  2HCl  →  2NaCl  +  H₂O  + CO₂

a) nNa₂CO₃ = 0,2.0,5 = 0,1 mol 

Theo tỉ lệ phản ứng => nHCl = 2nNa₂CO₃ = 0,1.2 = 0,2 mol

=> C_HCl = \(\dfrac{n}{V}\) = \(\dfrac{0,2}{0,2}\)=  1M

b) nNaCl = 2Na₂CO₃ = 0,2 mol

=> mNaCl = 0,2.58,5 = 11,7 gam

c) Thể tích dung dịch sau phản ứng = 0,2 + 0,2 = 0,4 lít

=> C_NaCl = \(\dfrac{0,2}{0,4}\)= 0,5M