Đáp án C

V ml K₂Cr₂O₇ 0,05M + 0,06 mol FeSO₄ + H₂SO₄
• Theo bảo toàn e: 6 × n_K2Cr2O7 = 1 × n_FeSO4 → n_K2Cr2O7 = 0,06 : 6 = 0,01 mol
→ V_K2Cr2O7 = 0,01 : 0,05 = 0,2 lít = 200 ml

a)\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

   \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

b)Gọi V là thể tích của X.

   \(n_Y=0,02\cdot0,25=5\cdot10^{-3}mol\)

   Để trung hòa X\(\Rightarrow n_X=n_Y=5\cdot10^{-3}mol\)

    \(\Rightarrow0,2\cdot V+0,1\cdot V=5\cdot10^{-3}\Rightarrow V=\dfrac{1}{60}l=16,67ml\)

Chọn C

a, \(m_{CH_3COOH}=100.6\%=6\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)

Theo PT: \(n_{NaHCO_3}=n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{NaHCO_3}=0,1.84=8,4\left(g\right)\)

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

b, Ta có: m _{dd sau pư} = 100 + 8,4 - 0,1.44 = 104 (g)

\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,1.82}{104}.100\%\approx7,88\%\)

\(n_{NaOH}=\dfrac{200.15\%}{40}=0,75\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,0001V\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,00005V\left(mol\right)\end{matrix}\right.\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

          0,0002V<-0,0001V

           6NaOH + Fe₂(SO₄)₃ --> 3Na₂SO₄ + 2Fe(OH)₃

         0,0003V<-0,00005V---------------->0,0001V

=> 0,0002V + 0,0003V = 0,75

=> V = 1500 (ml)

n_Fe(OH)3 = 0,15 (mol)

=> m₁ = 0,15.107 = 16,05 (g)

PTHH: 2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

              0,15--------->0,075

=> m_Fe2O3 = 0,075.160 = 12 (g)

\(a)H_2SO_4+Ba\left(NO_3\right)_2\rightarrow BaSO_4+2HNO_3\\ n_{H_2SO_4}=0,2.1=0,2l\\ n_{Ba\left(NO_3\right)_2}=n_{H_2SO_4}=0,2mol\\ m_{ddBa\left(NO_3\right)_2}=\dfrac{0,2.261}{20}\cdot100=261g\\ V_{ddBa\left(NO_3\right)_2}=\dfrac{261}{1,22}\approx213,9ml\\ c)n_{HNO_3}=0,2.4=0,4mol\\ C_{M_{HNO_3}}=\dfrac{0,4}{0,2+0,2139}\approx0,97M\)