Anh bổ sung câu c)

\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

Pt : \(2K+2H_2O\rightarrow2KOH+H_2|\)

       2           2               2          1

     0,2                         0,2

a) \(n_{KOH}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

\(C_{M_{ddKOH}}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

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

             1            1            1          1

           0,2          0,2

\(n_{HCl}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{ddHCl}=\dfrac{7,3.100}{15}\simeq48,67\left(g\right)\)

\(V_{ddHCl}=\dfrac{48,67}{1,2}=40,56\left(ml\right)\)

 Chúc bạn học tốt

\(n_{K_2O}=\dfrac{23.5}{94}=0.25\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

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

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

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

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

\(m_{dd_{H_2SO_4}}=\dfrac{0.25\cdot98}{20\%}=122.5\left(g\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{122.5}{1.14}=107.5\left(ml\right)=0.1075\left(l\right)\)

\(C_{M_{K_2SO_4}}=\dfrac{0.25}{0.1075+0.5}=0.4\left(M\right)\)

nK2O = 23,5 : 94 = 0,25 (mol)
Vdd = 500ml = 0,5l

PT K2O + H2O ==> 2KOH
TPT 1 1 2 (mol)
TĐB: 0,25 --> 0,5 (mol)

a) C_{M KOH} = 0,5 : 0,5 = 1(M)

b) PT: H2SO4 + 2KOH ==> K2SO4 + 2H2O
TPT: 1 2 1 2 (mol)

a.

\(n_{K_2O}=\dfrac{m}{M}=\dfrac{23,5}{94}=0,25\left(mol\right)\\ PT:\\ K_2O+H_2O\rightarrow2KOH\)

Theo pt: \(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)

\(\Rightarrow C_{M\left(KOH\right)}=\dfrac{n}{V}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b.

\(PT:\\ 2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)(2)

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

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{m_{dd}}{D}=\dfrac{n.M:C\%}{D}=\dfrac{0,25.98:20\%}{1,14}=\)107,46(ml)

c.

dd spư có chất tan là K₂SO₄

Theo pt (2), \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\)

=> C_M = \(\dfrac{n}{V}=\dfrac{0,25}{0,107}=2,336\left(M\right)\)

a)

$BaO + H_2O \to Ba(OH)_2$
$n_{Ba(OH)_2} = n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$C_{M_{Ba(OH)_2}} = \dfrac{0,2}{0,2} = 1M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,5} = 0,8(lít)$

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH (1)

a. Theo PT₍₁₎: \(n_{NaOH}=2.n_{Na_2O}=2.0,05=0,1\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,1}{2}=0,05M\)

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

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{4,9}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=24,5\left(g\right)\)

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

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

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

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

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

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: \(n_{OH^-}=n_{H^+}=0,5mol\)

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