PTPỨ : CO₂+Ca(OH)₂→CaCO₃+H₂O

a) 800ml = 0,8l

\(n_{H_2O}=\frac{m}{M}=\frac{2}{18}\approx0,1\) mol

n_CO2= C_M.V=0,1.0,8=0,08 mol

CO₂+Ca(OH)₂→CaCO₃+H₂O

So sánh tỉ lệ số mol của CO₂ với số mol của H₂O ta có:

\(\frac{0,08}{1}< \frac{0,1}{1}\) <=> n_CO2<n_H2O

=> CO₂ hết , H₂O dư

Theo ptpứ ta có : \(n_{H_2O\left(pứ\right)}=n_{CO_2}=0,08\) mol

=> n_CO2=0,1-0,08=0,02 mol

V_CO2(đktc)=n.22,4=0,02.22,4=0,448 l

b)

Theo ptpứ ta có : \(n_{CaCO_3}=n_{CO_2}=0,08\) mol

\(m_{CaCO_3}=n.M=0,08.100=8\)g

\(C_M=\frac{n}{V}=\frac{0,08}{V}\) hình như đề bạn thiếu pải k

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

0,02------0,02

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

0,08---------------0,08

a) n\(_{C_{ }a\left(OH\right)2}=0,8.0,1=0,08\left(mol\right)\)

n\(_{CaC_{ }O3}=\frac{2}{100}=0,02\left(mol\right)\)

Do sau phản ứng có 1 muối tan nên CO2 duư

Theo pthh1

n\(_{C_{ }O2}=n_{Ca\left(OH\right)2}=0,08\left(mol\right)\)

Theo pthh2

n\(_{CO2}=n_{CaCO3}=0,02\left(mol\right)\)

Σn\(_{CO2}=0,02+0,08=0,1\left(mol\right)\)

V\(_{CO2}=0,1.22,4=2,24\left(l\right)\)

b) Theo pthh2

n\(_{Ca\left(HCO3\right)2}=n_{CaC_{ }O3}=0,02\left(mol\right)\)

m\(_{Ca\left(HCO3\right)2}=0,02.162=3,24\left(g\right)\)

C_M Ca(HCO3)2=\(\frac{0,02}{0,8}=0,025\left(M\right)\)

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%

Bài 1 : 

$n_{CO_2} = \dfrac{3,136}{22,4} = 0,14(mol)$
$n_{Ca(OH)_2} = 0,8.0,1 = 0,08(mol)$

CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

0,08.......0,08...........0,08........................(mol)

CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂

0,06........0,06........................................(mol)

Suy ra : $m_{CaCO_3} = (0,08 - 0,06).100 = 2(gam)$

Bài 2 : 

$n_{CO_2} = \dfrac{2,24}{22,4} = 0,1(mol) ; n_{NaOH} = 0,1.1,5 = 0,15(mol)$
2NaOH + CO₂ → Na₂CO₃ + H₂O

0,15........0,075.......0,075....................(mol)

Na₂CO₃ + CO₂ + H₂O → 2NaHCO₃

0,025........0,025...................0,05..............(mol)

Suy ra:  

$C_{M_{NaHCO_3}} = \dfrac{0,05}{0,1} = 0,5M$
$C_{M_{Na_2CO_3}} = \dfrac{0,075 - 0,025}{0,1} = 0,5M$

b)

$NaOH + HCl \to NaCl + H_2O$
$n_{HCl} = n_{NaOH} = 0,15(mol)$
$m_{dd\ HCl} = \dfrac{0,15.36,5}{25\%} = 21,9(gam)$

a)

\(Fe + CuSO_4 \to FeSO_4 + Cu\)

Theo PTHH : 

\(n_{CuSO_4} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ \Rightarrow V_{dd\ CuSO_4} = \dfrac{0,2}{0,1} = 2(lít)\)

b)

\(n_{CuSO_4} = 0,1.2,5 = 0,25(mol) > n_{Fe}= 0,2\) nên \(CuSO_4\) dư.

Ta có : 

\(n_{CuSO_4\ pư} = n_{FeSO_4} = n_{Fe} = 0,2(mol)\\ \Rightarrow n_{CuSO_4\ dư} = 0,25 - 0,2 = 0,05(mol)\)

Vậy : 

\(C_{M_{CuSO_4}} = \dfrac{0,05}{2,5} = 0,02M\\ C_{M_{FeSO_4}} = \dfrac{0,2}{2,5} = 0,08M\)

\(n_{AgNO_3}=0,1.0,2=0,02\left(mol\right)\\ n_{HCl}=0,3.0,2=0,06\left(mol\right)\\ AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ a,Vì:\dfrac{0,02}{1}< \dfrac{0,06}{1}\Rightarrow HCldư\\ \Rightarrow n_{AgCl}=n_{HNO_3}=n_{AgNO_3}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,02=2,87\left(g\right)\\ b,dd.sau.p.ứ:HNO_3,HCl\left(dư\right)\\ n_{HCl\left(dư\right)}=0,06-0,02=0,04\left(mol\right)\\V_{ddsau}=V_{ddAgNO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ \Rightarrow C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,4}=0,1\left(M\right)\\ C_{MddHNO_3}=\dfrac{0,02}{0,4}=0,05\left(M\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ 2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\\ n_A=n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ n_B=n_{Cu}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ m_B=m_{Cu}=0,15.64=9,6\left(g\right)\\ C_{MddAlCl_3}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)