\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

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

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

\(1,C_{M\left(HCl\right)}=\dfrac{0,75}{0,5}=1,5M\\ 2,n_{Ca\left(OH\right)_2}=\dfrac{37}{74}=0,5\left(mol\right)\\ C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,5}{1,5}=0,33M\\ 3,n_{NaOH}=0,25+\dfrac{20}{40}=0,75\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,75}{2}=0,375M\\ 4,n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,5}{2}=0,25M\)

`1) C_[M_[HCl]] = [ 0,75 ] / [ 0,5 ] = 1,5 (M)`

_____________________________________________

`2)n_[Ca(OH)_2] = 37 / 74 = 0,5 (mol)`

`-> C_[M_[Ca(OH)_2]] = [ 0,5 ] / [ 1,5 ] ~~ 0,33 (M)`

_____________________________________________

`3) n_[NaOH] = 0,25 + 20 / 40 = 0,75 (mol)`

`-> C_[M_[NaOH]] = [ 0,75 ] / 2 = 0,375 (M)`

_____________________________________________

`4) n_[H_2 SO_4] = 49 / 98 = 0,5 (mol)`

`-> C_[M_[H_2 SO_4]] = [ 0,5 ] / 2 = 0,25 (M)`

\(n_{Ca\left(OH\right)_2}=0,2.0,4=0,08\left(mol\right)\)

PTHH: Ca(OH)₂ + 2CH₃COOH --> (CH₃COO)₂Ca + 2H₂O

              0,08---->0,16

=> \(C_{M\left(A\right)}=\dfrac{0,16}{0,3}=\dfrac{8}{15}M\)

\(n_{Ca\left(OH\right)_2}0,4.0,2=0,08\left(mol\right)\)

PTHH: 2CH₃COOH + Ca(OH)₂ ---> (CH₃COO)₂Ca + 2H₂O

            0,016<-------------0,08

\(\rightarrow C_{M\left(A\right)}=\dfrac{0,04}{0,3}=0,533M\)

400ml = 0,4l

\(n_{Ca\left(OH\right)2}=0,2.0,4=0,08\left(mol\right)\)

Pt : \(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O|\)

                 2                  1                         1                      2

               0,16              0,08

\(n_{CH3COOH}=\dfrac{0,08.2}{1}=0,16\left(mol\right)\)

300ml = 0,3l

\(C_{MddCH3COOH}=\dfrac{0,16}{0,3}=0,53\left(M\right)\)

 Chúc bạn học rốt

$n_{CO_2} = \dfrac{6,72}{22,4} = 0,3(mol) > n_{CaCO_3} = \dfrac{12}{100} = 0,12(mol)$

Do đó, có tạo muối axit

CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

0,12......0,12.............0,12..................(mol)

2CO₂ + Ca(OH)₂ → Ca(HCO₃)₂

0,18..........0,09................................(mol)

$n_{Ca(OH)_2} = 0,12 + 0,09 = 0,21(mol)$
$C_{M_{Ca(OH)_2}} = \dfrac{0,21}{0,1} = 2,1M$

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(n_{CaCO_3}=\dfrac{12}{100}=0,12\left(mol\right)\)

Vì tạo kết tủa nên CO2 phải phản ứng hết

=>Bảo toàn nguyên tố C : \(n_{Ca\left(HCO_3\right)_2}.2+n_{CaCO_3}=n_{CO_2}\)

=> \(n_{Ca\left(HCO_3\right)_2}=0,09\left(mol\right)\)

Bảo toàn nguyên tố Ca 

=> \(n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=n_{Ca\left(OH\right)_2}\)

=> \(n_{Ca\left(OH\right)_3}=0,21\left(mol\right)\)

=> \(CM_{\text{​​}Ca\left(OH\right)_2}=\dfrac{0,21}{0,1}=2,1M\)

nCaCO3=0,12(mol)

nCO2=0,3(mol)

PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O (1)

x___________x________x(mol)

CaCO3 + CO2 + H2O -> Ca(HCO3)2 (2)

y_________y_____________y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}x-y=0,12\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,21\\y=0,09\end{matrix}\right.\)

=> CMddCa(OH)2= 0,21/0,1=2,1(M)

a) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PTHH: \(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,1\left(mol\right)\)

\(V_{Ca\left(OH\right)_2}=200ml=0,2l\)

\(\Rightarrow C_{MCa\left(OH\right)_2}=\dfrac{n_{Ca\left(OH\right)_2}}{V_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)

b) Theo PTHH có: \(n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.74=7,4\left(g\right)\)

a) \(C\%=\dfrac{m_{KCl}}{m_{ddKCl}}.100\%=\dfrac{10}{300}.100\%\approx3,3\%\)

b) Đổi: \(1500ml=1,5l\)

\(C_{MCuSO_4}=\dfrac{n}{V}=\dfrac{3}{1,5}=2M\)

\(Ca\underrightarrow{1}CaO\underrightarrow{2}Ca\left(OH\right)_2\underrightarrow{3}CaCO_3\underrightarrow{4}CaO\)

(1) \(2Ca+O_2\underrightarrow{t^o}2CaO\)
(2) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

(3) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

(4) \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

 Chúc bạn học tốt

200ml = 0,2l

\(n_{Ca\left(OH\right)2}=0,1.0,2=0,02\left(mol\right)\)

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

              1             2              1             2

           0,02        0,04         0,02

a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)

300ml = 0,3l

\(C_{M_{ddHCl}}=\dfrac{0,04}{0,3}=0,13\left(M\right)\)

b) \(n_{CaCl2}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)

\(V_{ddspu}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{CaCl2}}=\dfrac{0,02}{0,5}=0,04\left(M\right)\)

 Chúc bạn học tốt