a) \(n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: K₂O + H₂O --> 2KOH

              0,4------------>0,8

=> m_KOH = 0,8.56 = 44,8 (g)

b) \(C_M=\dfrac{0,8}{0,5}=1,6M\)

c) 

PTHH: 4K + O₂ --t^o--> 2K₂O

                     0,2<----0,4

=> V_O2 = 0,2.22,4 = 4,48 (l)

nKK 2O = 37,6\94 = 0,4 mol

K2O+H2O --> 2KOH

 0,4                0,4                   0,8        ( mol )

mKOH= 0,8.56=44,8g

CMMKOH = 0,80,50,80,5 = 1,6M

c. 

4K+O2 --to> 2K2O

0,4       0,1                                 ( mol )

VO2 = 0,1.22,4=2,24lít

\(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\\ PTHH:K_2O+H_2O\rightarrow2KOH\\ \Rightarrow n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\\ \Rightarrow C_{M_{KOH}}=\dfrac{0,5}{0,2}=2,5M\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,25\cdot98=24,5\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{24,5\cdot100\%}{49\%}=50\left(g\right)\)

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

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

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

đề bài ko rõ lắm, 0,51 hay là sao

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

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25                       0,5

\(C_{M_{NaOH}}=\dfrac{0,5}{0,51}=0,98M\)

Để trung hòa: \(n_{OH^-}=n_{H^+}=0,5mol\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=\dfrac{1}{2}\cdot0,5=0,25mol\)

\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

         1           1              2

        0,1                        0,2

a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddKOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)

              1            1           1            1

            0,1          0,1

\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)

        1          2               1              1

      0,05      0,1

\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

 Chúc bạn học tốt

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

a) PTHH: Na2O + H20 -> 2NaOH
số mol Na20 = 0,25 (mol)
=> số mol NaOH = 0,5 mol.
Nôngd độ mol NaOH = 0,5 / 0,5 = 1 M
b) PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O
số mol H2SO4 = 1/2 số mol NaOH = 0,25 mol
C% H2SO4 = mH2SO4 / m ddH2SO4 . 100%
=> m ddH2SO4= 122,5 g
D=m/V => V= 107,5 ml

Số mol Na₂O = 15,5:62 = 0,25 mol

a) Khi cho Na₂O xảy ra phản ứng, tạo thành phản ứng dung dịch có chất tan là NaOH.

Na₂O + H₂O → 2NaOH

Phản ứng: 0,25 → 0,05 (mol)

500 ml = = 0,5 lít; C_{M, NaOH} = = 1M.

b) Phương trình phản ứng trung hòa dung dịch:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Phản ứng: 0, 5 → 0,25 0,25 (mol)

m_H2SO4 = 0,25x98 = 24,5 g

m_{dd H2SO4} = = 122,5 g

m_{dd, ml} = = ≈ 107,5 ml

nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

nNa2O = 0,125 mol

a. Na2O + H2O --------> NaOH

0,125 mol ----------------> 0,125 mol

--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M

b. H2SO4 + 2NaOH ------> Na2SO4 + H2O

....0,0625 <---0,125 mol

--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g

--> mH2SO4(20%) = 6,125/20% = 30,625 g

suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

nNa2O = 0,125 mol

a. Na2O + H2O --------> NaOH

0,125 mol ----------------> 0,125 mol

--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M

b. H2SO4 + 2NaOH ------> Na2SO4 + H2O

....0,0625 <---0,125 mol

--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g

--> mH2SO4(20%) = 6,125/20% = 30,625 g

suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

a, \(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)

PTHH: K₂O + H₂O → 2KOH

Mol:     0,05                   0,1

b) \(C_{M_{ddKOH}}=\dfrac{0,1}{0,02}=5M\)

c) 

PTHH: KOH + HCl → KCl + H₂O

Mol:      0,1        0,1      0,1

\(m_{ddHCl}=\dfrac{0,1.36,5.100}{20}=18,25\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{18,25}{0,9125}=103,9\left(ml\right)=0,1039\left(l\right)\)

d) \(C_{M_{ddKCl}}=\dfrac{0,1}{0,02+0,1039}=0,8071M\)