\(n_{H_2SO_4}=0,2.0,25=0,05 mol \)
a) \(H_2SO_4+2KOH \rightarrow K_2SO_4+2H_2O\)
Theo PTHH ta có: \(n_{KOH}=2n_{H_2SO_4}=0,05.2=0,1 mol\)
\(\rightarrow V_{dd KOH}=\frac{n}{C_M}=\frac{0,1}{0,5}=0,2 \left(l\right)=200\left(ml\right)\)
\(n_{K_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\)
\(m_{dd sau}=250+200=450=0,45\left(l\right)\)
\(\rightarrow C_{M K_2SO_4}=\frac{0,05}{0,45}=\frac{1}{9}M\)
b)\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{NaOH}=2.n_{H_2SO_4}=0,1\left(mol\right)\rightarrow m_{NaOH}=0,1.40=4\left(g\right)\)
\(m_{dd NaOH}=\frac{4.100}{20}=20\left(g\right)\)
\(n_{Na_2SO_4}=0,05\left(mol\right)\rightarrow m_{Na_2SO_4}=0,05.142=7,1\left(g\right)\)
Do D=1ml/g -> mdd H2SO4=250(g)
\(C\%_{Na_2SO_4}=\frac{7,1}{250+20}.100=2,63\%\)

1. Ptrình ion H(+) + OH(-) = H2O 
n H(+) 0,3*0,75*2 + 0,3*1,5 = 0,9mol 
=> n OH(-) = 0,9mol => n KOH = 0,9mol => V = 0,6l

2. a) nNaOH= 0,05.20/40=0,025 mol 
NaOH + HCl ------> NaCl +H2O 
....3x.........3x 
2NaOH +H2SO4------> Na2SO4 + 2H2O 
.....2x.........x 
tỉ lệ mol 2 axit HCl : H2SO4 =3:1 
đặt số mol H2SO4 la` x ----> nHCl =3x 
>>>>3x+2x =0,025 >>>x=0,05 mol 
=>nồng độ mol của HCl va` H2SO4 lần lươt la` 1,5M & 0,5M 

b) n(OH-) = nNaOH + 2nBa(OH)2 = 0,2V + 2.0,1.V=0,4V 
trong 0,2l ddA có 0,3 mol HCl & 0,1 mol H2SO4 ( vi` V gấp đôi >> n gấp đôi) 
=> n(H+)= nHCl + 2nH2SO4 = 0,5mol 
ma` n(OH-) =n(H+) 
=> 0,4V=0,5 >>V= 1,25l=1250ml 

c) nNaOH=0,2.1,25=0,25mol = nBa(OH)2 
nH2O = n(axit)= 0,3 +0,1 =0,4 mol 
theo BTKL : m(muối) = m(axit) + m(bazo) -m(H2O) 
..............................= 0,3.36,5 +0,1.98 + 0,25( 40+171) -0,4.18=66,3g 

a) H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

b) \(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2NaOH --> Na₂SO₄ + 2H₂O

                 0,2---->0,4

=> m_NaOH = 0,4.40 = 16 (g)

=> \(m_{dd.NaOH}=\dfrac{16.100}{20}=80\left(g\right)\)

c)

PTHH: H₂SO₄ + 2KOH --> K₂SO₄ + 2H₂O

                0,2---->0,4

=> m_KOH = 0,4.56 = 22,4 (g)

=> \(m_{dd.KOH}=\dfrac{22,4.100}{5,6}=400\left(g\right)\)

=> \(V_{dd.KOH}=\dfrac{400}{1,045}=382,775\left(ml\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ pthh:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) 
          0,2                   0,4 
\(m_{\text{ }NaOH}=0,4.40=16g\\ m_{\text{dd}NaOH}=\dfrac{16.100}{20}=80g\) 
\(pthh:H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\) 
            0,2             0,4 
\(m_{KOH}=0,4.56=22,4g\\ m_{\text{dd}KOH}=\dfrac{22,4.100}{5,6}=400g\\ V_{\text{dd}}=\dfrac{400}{1,045}=382,7ml\)
\(V_{dd}=\dfrac{22,4}{1,045}=21,4354ml\)

a)PTHH:
HCl + NaOH → NaCl + H2O
nHCl = 0,04 (mol) = nNaOH = nNaCl
=>VddNaOH = 0,04/0,1 = 0,4 (l) = 400 (ml)
Vdd = VddNaOH + VddHCl = 0,6 (l)
=>C(M) ≈ 0,067 (M)
b) 2HCl + Ca(OH)2 → CaCl2 + 2H2O
nCa(OH)2 = nCaCl2 = (1/2)nHCl = 0,02 (mol)
(Nồng độ phần trăm = 25% ????)
mCa(OH)2 = 1,48 (g)
=>mdd(Ca(OH)2) = 5,92 (g)
mddHCl = 220 (g)
=>mdd = 225,92 (g)
mCaCl2 = 2,22 (g)
=>%mCaCl2 ≈ 0,98%

a.250ml=0,25l ; nHCl=0,25.1,5=0,375mol

KOH+HCl->KCl+H2O

1mol 1mol 1mol

0,375 0,375 0,375

VKOh=0,375/2=0,1875l

b.CM KCL=0,375/0,25=1,5M

c.NaOH+HCL=NaCl+H2O

1mol 1mol

0,375 0,375

mdd NaOH=0,375.40.100/10=150g

n_HCl = C_M . V = 1,5 . 0,25 = 0,375 mol

PTHH: KOH + HCl -------> KCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <-0,375 -----> 0,375-> 0,375 (mol)

a) V_KOH = \(\dfrac{n}{C_M}\)= 0,1875 l

b) C_M KCl = \(\dfrac{n}{V}\)= \(\dfrac{0,375}{0,25+0,1875}\)\(\approx\)0,86 M

c) PTHH: NaOH + HCL ----> NaCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <- 0,375 ------>0,375 --> 0,375 (mol)

m_ctNaOH = n . M = 15 g

m_{ddNaOH = \(\dfrac{15.100\%}{10\%}\)=} 150 g

:3

\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\\ n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

a

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

0,02<------0,04----->0,02

Xét \(\dfrac{0,3}{1}>\dfrac{0,04}{2}\Rightarrow Ca\left(OH\right)_2.dư\)

\(m_{CaCl_2}=0,02.111=2,22\left(g\right)\)

b

Muốn pứ xảy ra hoàn toàn phải thêm dung dịch HCl 0,2 M

\(n_{HCl.cần}=2n_{Ca\left(OH\right)_2}=0,3.2=0,6\left(mol\right)\\ n_{HCl.cần.thêm}=0,6-0,04=0,56\left(mol\right)\)

\(V_{cần.\left(HCl\right)}=\dfrac{0,56}{0,2}=2,8\left(l\right)=280\left(ml\right)\\ V_{cần.thêm\left(HCl\right)}=280-200=80\left(ml\right)\)

c

\(CM_{CaCl_2}=\dfrac{0,02}{0,3+0,28}=\dfrac{1}{29}M\)

\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\\ n_{HCl}=0,2.0,2=0,04\left(mol\right)\\ Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

  0,02<----0,04------>0,02

a

\(m_{CaCl_2}=0,02.111=2,22\left(g\right)\)

b

Muốn pứ xảy ra hoàn toàn thì phải thêm dd HCl 0,2M.

\(n_{HCl.cần}=0,3.2=0,6\left(mol\right)\\ n_{HCl.cần.thêm}=0,6-0,04=0,56\left(mol\right)\\ V_{HCl.cần.}=\dfrac{0,56}{0,2}=2,8\left(l\right)\\ V_{HCl.cần.thêm}=2,8-0,2=2,6\left(l\right)=260\left(ml\right)\)

c

\(CM_{CaCl_2}=\dfrac{0,3}{0,3+2,8}=\dfrac{3}{31}M\)

200 ml dung dịch gì