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, \(Na_2O+H_2O\rightarrow2NaOH\)

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

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

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

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

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

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

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

a)

$Na_2O + H_2O \to 2NaOH$

n Na2O = 15,5/62 = 0,25(mol)

n NaOH = 2n Na2O = 0,5(mol)

=> CM NaOH = 0,5/0,5 = 1M

b) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

n H2SO4 = 1/2 n NaOH = 0,25(mol)

=> m dd H2SO4 = 0,25.98/20% = 122,5(gam)

=> V dd H2SO4 = m / D = 122,5/1,14 =107,46(ml)

c) n Na2SO4 = n H2SO4 = 0,25(mol)

CM Na2SO4 = 0,25/0,10746 = 2,33M

a)

`\(Na_2O++H_{2_{ }}O->2NaOH\)

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

\(n_{Na_2O}=2n_{Na_2O}=0,5mol\)

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

\(n_{Na_2O}=\dfrac{15.5}{62}=0.25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(0.25.......................0.5\)

\(C_{M_{NaOH}}=\dfrac{0.5}{0.5}=1\left(M\right)\)

b.

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

\(0.5..............0.25\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.25}{2}=0.125\left(l\right)\)

\(a.n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 0,25...................................0,5\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ 0,25............0,5..........0,25\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,25}{2}=0,125\left(l\right)\)

BaO+H2O -> Ba(OH)2 
0,02             0,02  
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
      0,02          0,02
=> m = 0,392 g 
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l

Na2O + H2O -> 2NaOH

 0.1                       0.2

a.\(nNa2O=\dfrac{6.2}{62}=0.1mol\)

\(CM_{NaOH}=\dfrac{0.2}{0.3}=0.6M\)

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

     0.2            0.1

\(mH2SO4=0.1\times98=9.8g\)

\(mddH2SO4=\dfrac{9.8\times100}{20}=49g\)

\(V_{H2SO4}=\dfrac{49}{1.14}=43ml=0.043l\)

K₂O + H₂O -> 2KOH (1)

n_K2O=\(\dfrac{14,1}{94}=0,15\left(mol\right)\)

Theo PTHH 1 ta có:

2n_K2O=n_KOH=0,3(mol)

C_M dd KOH=\(\dfrac{0,3}{0,5}=0,6M\)

c;

2NaOH + H₂SO₄ -> Na₂SO₄ + 2H₂O (2)

Theo PTHH 2 ta có:

\(\dfrac{1}{2}\)n_NaOH=n_H2SO4=0,15(mol)

m_H2SO4=98.0,15=14,7(g)

m_{dd H2SO4}=14,7:10%=147(g)

V_{dd H2SO4}=147:1,14=129(ml)

a, K₂O + H₂O ->2 KOH

b, n_K2O= 0,15 ( mol )

K2O + H2O-> 2KOH

Theo pt 1 1 2 ( mol )

Theo đb 0,15 0,15 0,3 ( mol)

==> C_M=\(\dfrac{0.3}{0,5}\) = 0,6 M

c, 2KOH + H₂SO₄ -> K₂SO₄ + 2H₂O

Theo pt 2 1

Theo đb 0,15 0,075

==> m_{dd H2S04}=\(\dfrac{0,075.98.100\%}{10\%}=73,5\left(g\right)\)

==> V_{dd h2so4}=\(\dfrac{73,5}{1,14}=64,47\left(ml\right)\)

nhầm câu c tí nha NaOH -> KOH nha ko có ảnh hưởng gi đến bài giải đâu