d

D

Đáp án là B. 80%

Đáp án D

\(n_{AgCl}=\dfrac{7,175}{143,5}=0,05\left(mol\right)\)

PTHH: HCl + AgNO₃ ---> AgCl↓ + HNO₃

           0,05<---------------0,05
\(\rightarrow m_{HCl}=0,05.36,5=1,825\left(g\right)\\ \rightarrow C\%_{ddA}=\dfrac{1,825}{50}.100\%=3,65\%\)

\(n_{Cl_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Đặt H = x%

PTHH: Cl₂ + H₂ --as--> 2HCl

LTL: 6,72 < 10 => H₂ dư

=> n_HCl = 0,3x (mol)

\(\rightarrow C\%_{HCl}=\dfrac{0,3x.36,5}{0,3x.36,5+385,4}.100\%=3,65\%\\ \Leftrightarrow20,23\%\)

\(n_{H_2}=\dfrac{1}{22,4}=\dfrac{5}{112}\left(mol\right)\)

\(n_{Cl_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: H₂ + Cl₂ --as--> 2HCl

Xét tỉ lệ: \(\dfrac{\dfrac{5}{112}}{1}>\dfrac{0,03}{1}\) => Hiệu suất tính theo Cl₂

\(n_{AgCl}=\dfrac{7,175}{143,5}=0,05\left(mol\right)\)

Bảo toàn Cl: \(n_{Cl_2\left(pư\right)}=0,025\left(mol\right)\)

\(H\%=\dfrac{0,025}{0,03}.100\%=83,33\%\)

PTHH :

\(H_2+Cl_2\rightarrow2HCl\left(1\right)\)

\(HCl+AgNO_3\rightarrow AgCl+HNO_3\) \(\left(2\right)\)

Ta có : \(n_{H_2}=\frac{V}{22,4}=\frac{1}{22,4}=0,446\left(mol\right)\)

\(n_{Cl_2}=\frac{V}{22,4}=\frac{0,672}{22,4}=0,3\left(mol\right)\)

Vì \(n_{H_2}>n_{Cl_2}\Leftrightarrow n_{H_2}\) dư, \(n_{Cl_2}\) hết

Theo pt (2) : \(n_{HCl}=n_{AgCl}=0,01\left(mol\right)\) (trong 5g)

\(\Leftrightarrow n_{HCl}=0,01.4=0,04\left(mol\right)\) (trong 20g)

Theo pt (1) : \(n_{Cl_2pư}=\frac{1}{2}n_{HCl}=0,02\left(mol\right)\)

\(H=\frac{0,02}{0,03}.100\%=66,67\%\)

Vậy....

nH2=\(\frac{12}{2,4}\)=0,0446mol

nCl2=0,03mol

H₂ + Cl₂ → 2HCl (1)

Vì nH2>nCl2nH2>nCl2\(\rightarrow\) Hiệu suất phản ứng tính theo Cl₂

AgNO₃ + HCl → AgCl + HNO₃
\(\rightarrow\) n_{HCl (trong 5g A)} = n_AgCl = 0,01 mol
\(\rightarrow\) n_{HCl (trong 20g A)} = 0,01.4 = 0,04 mol
(1) => n_{Cl2 phản ứng} = 0,5.n_HCl = 0,5.0,04 = 0,02 mol
=> H% = \(\frac{0,02}{0,03}.100\%\)= 66,67%

\(H_2+Cl_2\rightarrow2HCl\)

\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)

\(n_{H2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)

\(n_{CL2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)

\(\rightarrow\) Tính theo Cl2

\(n_{AgCl}=\frac{1,435}{143,5}=0,01\left(mol\right)\)

\(\rightarrow n_{HCl_{bđ}}=0,02\left(mol\right)\)

\(\rightarrow n_{HCl_{lt}}=0,025.2=0,05\left(mol\right)\)

\(\rightarrow H=\frac{0,02}{0,05}.100=40\%\)

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

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

=> 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