1)

- TN1:

\(n_{AgCl}=\dfrac{35,875}{143,5}=0,25\left(mol\right)\)

PTHH: AgNO₃ + HCl --> AgCl + HNO₃

                          0,25<--0,25

TN2:

n_NaOH = 0,5.0,3 = 0,15 (mol)

PTHH: NaOH + HCl --> NaCl + H₂O

             0,15--->0,15

\(n_{HCl\left(dd.C\right)}=0,25+0,15\) = 0,4 (mol)

=> \(C_{M\left(dd.C\right)}=\dfrac{0,4}{2}=0,2M\)

2)

Có \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{V}M\\C_{M\left(B\right)}=\dfrac{0,15}{V^,}M\end{matrix}\right.\)

n_HCl(A) = \(\dfrac{0,025}{V}\left(mol\right)\) 

PTHH: Fe + 2HCl --> FeCl₂ + H₂

                     \(\dfrac{0,025}{V}\)------>\(\dfrac{0,0125}{V}\)

n_HCl(B) = \(\dfrac{0,015}{V^,}\) (mol)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

                    \(\dfrac{0,015}{V^,}\)-------->\(\dfrac{0,0075}{V^,}\)

TH1: \(\dfrac{0,0125}{V}=\dfrac{0,0075}{V^,}+0,02\) 

Mà V + V^' = 2 (l)

=> \(\left[{}\begin{matrix}V=1,5;V^,=0,5\left(KTM\right)\\V=0,5;V^,=1,5\left(TM\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{0,5}=0,5M\\C_{M\left(B\right)}=\dfrac{0,15}{1,5}=0,1M\end{matrix}\right.\)

TH2: \(\dfrac{0,0125}{V}+0,02=\dfrac{0,0075}{V^,}\)

=> \(\left[{}\begin{matrix}V=\dfrac{1+\sqrt{6}}{2};V^,=\dfrac{3-\sqrt{6}}{2}\\V=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{\dfrac{1+\sqrt{6}}{2}}=\dfrac{-1+\sqrt{6}}{10}M\\C_{M\left(B\right)}=\dfrac{0,15}{\dfrac{3-\sqrt{6}}{2}}=\dfrac{3+\sqrt{6}}{10}M\end{matrix}\right.\)

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Đáp án A

n_OH-= n_H+ nên 0,02.a+ 0,02.b = 0,3.0,1

n_Cl-= n_AgCl = 0,02 mol = 0,02a suy ra a = 1M suy ra b = 0,5

Đáp án A

Đáp án A

Đáp án A

Ta có: n B a C O 3 = 11,82/197 =  0,06 mol; n C a C O 3 = 7/100 = 0,07 mol

NaOH + NaHCO₃  Na₂CO₃ + H₂O  (1)

    0,06                           0,06 mol

    BaCl₂ + Na₂CO₃ BaCO₃ + 2NaCl  (2)

                  0,06             0,06 mol

    m = 0,06.2.40 = 4,8g

    2NaHCO₃  Na₂CO₃ + CO₂ + H₂O  (3)

    CaCl₂ + Na₂CO₃ CaCO₃ + 2NaCl  (4)

                   0,07           0,07 mol

    n N a 2 C O 3 ₍₄₎ =  n N a 2 C O 3  ₍₁₎ +  n N a 2 C O 3  ₍₃₎

→ n N a 2 C O 3  ₍₃) = 0,07 – 0,06 = 0,01 mol

n N a H C O 3 trong một lít dd =  n N a H C O 3  ₍₁₎ +  n N a H C O 3  ₍₃₎ = 0,06 + 0,01.2 = 0,08 mol

a = 0,08/1 = 0,08 mol/l

a. HCl + AgNO3 --> AgCl + HNO3;

0,02--------------------0,02 (mol);

H2SO4 + BaCl2---> BaSO4 + 2HCl;

0,02-----------------------0,02 (mol);

ta có: n_AgCl= 2,87/143,5= 0,02(mol); n_BaSO4= 4,66/233=0,02(mol);

=> n_{HCl= 0,02(mol);} n_BaSO4=0,02 (mol)

C_{M HCl}= C_{M BaSO4}= 0,02/0,05=0,4M

b. HCl + NaOH --> NaCl + H2O;

0,02--------0,02 (mol);

H2SO4 + 2NaOH --> Na2SO4 +2 H2O;

0,02----------0,04 (mol);

ta có: n_{NaOH cần} = 0,02+0,04=0,06 (mol);

=> V_{NaOH cần} = 0,06/0,2=0,3(l) = 300ml

Đáp án A

P1: n_Fe(OH)2 = 3,6/90 = 0,04 mol => n_Fe2+ = 0,04 mol

P2:

Fe²⁺: x mol

H⁺: y mol

Cl^-: 2x + y

3Fe²⁺ + 4H⁺ + NO₃^- → 3Fe³⁺ + NO + 2H₂O

0,06 ← 0,08          ←                0,02

=> y = 0,08

Fe²⁺ + Ag⁺ → Fe³⁺ + Ag

x-0,06        →      x-0,06

Ag⁺ + Cl^- → AgCl

         2x+y → 2x+y

=> 108(x-0,06) + 143,5(2x+y) = 68,2 => x = 0,16

=> n_{Fe phản ứng ban đầu} = n_H2 = 0,04+0,16 = 0,2 mol => V_H2 = 4,48 lít