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%

Theo đề bài ta có : nHCl = 0,2.0,2=0,04(mol)

a) Ta có PTHH :

\(HCl+NaOH\rightarrow NaCl+H2O\)

0,04mol.....0,04mol....0,04mol

Ta có :

\(V_{\text{dd}HCl\left(c\text{ần}-d\text{ùng}\right)}=\dfrac{0,04}{0,1}=0,4\left(lit\right)=400\left(ml\right)\)

CM_NaCl = \(\dfrac{0,04}{0,2}=0,2\left(M\right)\)

b) Theo đề bài ta có : mddHCl=\(200.1=200\left(g\right)\)

Ta có PTHH :

\(Ca\left(OH\right)2+2HCl\rightarrow CaCl2+2H2O\)

0,02mol...........0,04mol....0,02mol

Ta có :

\(m\text{dd}Ca\left(OH\right)2\left(c\text{ần}-d\text{ùng}\right)=\dfrac{0,02.74}{5}.100=29,6\left(g\right)\)

C%_CaCl2 = \(\dfrac{0,02.111}{0,02.74+200}.100\%\approx1,102\%\)

Vậy..............

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

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

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

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 
\(m_{KOH}=0,4.56=22,4g\\ m_{\text{dd}KOH}=\dfrac{22,4.100}{5,6}=400g\\ V_{\text{dd}}=\dfrac{400}{1,045}=382,7ml\)
\(V_{dd}=\dfrac{22,4}{1,045}=21,4354ml\)

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 

Theo đề bài ta có :

VddHCl=200ml=0,2 l

nHCl = 0,2.0,2=0,04 mol

a) Ta có PTHH 1:

HCl + NaOH \(\rightarrow\) NaCl + H2O

0,04mol...0,04mol....0,04mol

=> \(\left\{{}\begin{matrix}V\text{dd}NaOH=\dfrac{0,04}{0,1}=0,4\left(l\right)=400\left(ml\right)\\CM_{\text{dd}NaCl\left(c\text{ần}-d\text{ùng}\right)}=\dfrac{0,04}{0,2}=0,2\left(M\right)\end{matrix}\right.\)

b) Ta có PTHH 2 :

2HCl + Ca(OH)2 -> CaCl2 + 2H2O

0,04mol...0,02mol.....0,02mol

Ta có : mdd(sau-p/ư) = 200.1 + 0,02.74 = 201,48 g

=> C%_CaCl2 = \(\dfrac{\left(0,02.111\right)}{201,48}.100\%\approx1,102\%\)

Vậy....

\(n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,04mol <---0,04mol

\(V_{NaOH}=\dfrac{0,04}{0,1}=0,4\left(l\right)=400\left(ml\right)\)

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

0,02mol<--- 0,04mol

\(m_{dd_{Ca\left(OH\right)_2}}=\dfrac{0,02.74.100}{6}=24,67\left(g\right)\)