Anh bổ sung câu c)

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

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)

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

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

\(K_2O+H_2O\rightarrow2KOH\)

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

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

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

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

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

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

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

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

\(n_{NaOH}=0,5.0,2=0,1\left(mol\right);n_{HCl}=0,5.0,3=0,15\left(mol\right)\)

PTHH: NaOH + HCl → NaCl + H₂O

Mol:        0,1      0,1         0,1

Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ NaOH hết, HCl dư

V_{dd sau pứ} = 0,2 + 0,3 = 0,5 (l)

\(C_{M_{ddNaCl}}=\dfrac{0,1}{0,5}=0,2M\)

\(C_{M_{ddHCldư}}=\dfrac{0,15-0,1}{0,5}=0,1M\)

Bài1:

a,Vì dd A là dd bazo nên làm cho quỳ tím đổi thành màu xanh

b,\(n_{Na_2O}=\dfrac{21,7}{62}=0,35\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:       0,35                     0,7

\(\Rightarrow C_{M_{ddNaOH}}=\dfrac{0,7}{0,4}=1,75M\)

Bài 2:

a,\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,15     0,3                      0,15

⇒ a=m_Zn = 0,15.65 = 9,75 (g)

b,\(V_{HCl}=\dfrac{0,3}{1,5}=0,2\left(l\right)=200\left(ml\right)\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ a,C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{ddH_2SO_4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\\ V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,456\left(ml\right)\\ c,V_{ddsau}=V_{ddNaOH}+V_{ddH_2SO_4}\approx0,5+0,107456=0,607456\left(l\right)\\C_{MddNa_2SO_4}\approx\dfrac{ 0,25}{0,607456}\approx0,411552\left(M\right)\)

a. \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(C_M=\dfrac{0,1}{200}=0,0005\left(mol\text{/}l\right)\) 

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M)