Lần sau bạn đăng tách từng bài ra nhé.

Câu 1:

a, \(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

Câu 3: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

\(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)

Câu 4: \(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PT: \(CuSO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+CuCl_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\), ta được CuSO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

Câu 2:

\(n_{HCl}=0,3.0,2=0,06\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,06\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,06}{1,5}=0,04\left(l\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,06}{0,3+0,04}=\dfrac{3}{17}\left(M\right)\)

\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)

a, \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

a, \(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=0,25.2=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)

\(n_{BaO}=\dfrac{30,6}{153}=0,2mol\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

 0,2                    0,2

 Để trung hòa: \(n_{OH^-}=n_{H^+}=0,2mol\)

 \(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)

 \(\Rightarrow m_{ddHCl}=\dfrac{7,3}{14,6}\cdot100=50\left(g\right)\)

PTHH: BaO + H₂O ---> Ba(OH)₂ (1)

Ba(OH)₂ + 2HCl ---> BaCl₂ + 2H₂O (2)

Ta có: \(n_{BaO}=\dfrac{30,6}{153}=0,2\left(mol\right)\) 

Theo PT₍₁₎: \(n_{Ba\left(OH\right)_2}=n_{BaO}=0,2\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{Ba\left(OH\right)_2}=2.0,2=0,4\left(mol\right)\)

=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{14,6}{m_{dd_{HCl}}}.100\%=14,6\%\)

=> \(m_{dd_{HCl}}=100\left(g\right)\)

1. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:      0,25                     0,5

\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)

2. 

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:        0,5           0,25

\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)

\(V_{ddH_2SO_4}=\dfrac{122,5}{1,14}=107,456\left(ml\right)\)

\(n_{Na2O}=\dfrac{m_{Na2O}}{M_{Na2O}}=0,25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25 mol - 0,25 mol - 0,5 mol

a) \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{NaOH}}=1\left(M\right)\)

b) \(H_2SO_4+2NaOH\rightarrow Na_2SO4+2H_2O\)

0,25 mol - 0,5 mol - 0,25 mol - 0,5 mol

\(m_{ctH2SO4}=n_{H2SO4}.M_{H2SO4}=24,5\left(g\right)\)

\(C_{\%_{H2SO4}}=\dfrac{m_{ctH2SO4}}{m_{ddH2SO4}}.100\%\)

\(\Rightarrow m_{ddH2SO4}=\dfrac{m_{ctH2SO4}.100\%}{C_{\%_{H2SO4}}}=122,5\left(g\right)\)

\(D_{H2SO4}=\dfrac{m_{ddH2SO4}}{V_{H2SO4}}\Rightarrow V_{H2SO4}=\dfrac{m_{ddH2SO4}}{D_{H2SO4}}\approx107,46\left(ml\right)\)

Chọn A

a, \(n_{Na_2O}=\dfrac{7,75}{62}=0,125\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:     0,125                    0,25

b, \(C_{M_{ddNaOH}}=\dfrac{0,25}{0,25}=1M\)

c, 

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:        0,25       0,125

\(m_{ddH_2SO_4}=\dfrac{0,125.98.100}{20}=61,25\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{61,25}{1,14}=53,728\left(ml\right)\)