a)

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$

b)

$n_{CO_2} = n_{CaCO_3} = \dfrac{20}{100} = 0,2(mol)$

$V_{CO_2} = 0,2.22,4 = 4,48(lít)$

c)

Sau phản ứng : 

$m_{dd} = 20 + 200 - 0,2.44 = 211,2(gam)$

$n_{CaCl_2} = n_{CaCO_3} = 0,2(mol)$

$C\%_{CaCl_2} = \dfrac{0,2.111}{211,2}.100\% = 10,51\%$

a, PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\text{​​}\right)\)

\(\Rightarrow V_{CO_2}=0,24,48\left(l\right)\)

c, \(m_{ddsaupu}=200+20-44.0,2=211,2\left(g\right)\)

\(\Rightarrow C\%=\dfrac{0,2.111}{211,2}.100\%=10,51\%\)

\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ a.CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b.n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,2\left(mol\right)\\ m_{ddCaCl_2}=20+200-0,2.44=211,2\left(g\right)\\ V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c.m_{CaCl_2}=0,2.111=22,2\left(g\right)\\ C\%_{ddCaCl_2}=\dfrac{22,2}{211,2}.100\approx10,511\%\)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\)

c, \(n_{CH_3COOH}=2n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{60\%}=20\left(g\right)\)

n_CaCO3=20/100=0,2 mol

PTHH: CaCO₃ +2HCl--> CaCl₂ +CO₂+H₂O

--> n_CO2=n_CaCO3=0,2 mol;

--> V_CO2=0,2*22,4=5,6l;

ta có n_HCl= 2n_CaCO3= 2*0,2 =0,4 mol

--> m_{HCl phản ứng}= 0,4*36,5=14,6(g);

--> m_{dd HCl}=14,6/7,3%=200(g);

--> Khối lương dd sau phản ứng là 200+ 20-0,2*44=211,2(g);

C%_CaCl2=0,2*111/211,2=10,51%

nCaCO3= 20/100=0.2 mol

CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

0.2_______0.4_____0.2_____0.2

mHCl= 0.4*36.5=14.6g

mddHCl= 14.6*100/7.3=200g

mCO2= 0.2*44=8.8g

VCO2= 0.2*22.4=4.48l

mdd sau phản ứng= mCaCO3+mddHCl - mCO2= 20+200-8.8=211.2g

mCaCl2= 0.2*111=22.2g

C%CaCl2= 22.2/211.2*100%= 10.51%

a) 

\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

               0,15<---------0,3<------------------------0,15

=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)

b)

\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)

c)

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

                    0,3--------->0,15

=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)

a, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

PTHH: 2CH₃COOH + CaCO₃ ---> (CH₃COO)₂Ca + CO₂↑ + H₂O

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

b, \(\left\{{}\begin{matrix}V_1=\dfrac{0,2}{0,01}=20\left(l\right)\\V_2=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

c, \(n_{CO_2}=0,1.75\%=0,075\left(mol\right)\)

PTHH: Ca(OH)₂ + CO₂ ---> CaCO₃↓ + H₂O

                           0,075--->0,075

=> m_giảm = m_CaCO3 - m_CO2 = 0,075.100 - 0,075,44 = 4,2 (g)

\(n_{CaCO_3}=\dfrac{10}{100}=0,1mol\)

a)\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

       0,2                   0,1                  0,1                   0,1      0,1

b)\(V_1=V_{CH_3COOH}=\dfrac{0,2}{0,01}=20l\)

   \(V_2=0,1\cdot22,4=2,24l\)

c)Dẫn 75% khí \(CO_2\) qua dung dịch \(Ca\left(OH\right)_2\) thì: \(n_{CO_2}=0,1\cdot75\%=0,075mol\)

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

   0,075    0,075           0,075

   Khối lượng \(Ca\left(OH\right)_2\) giảm và giảm:

   \(\Delta m=m_{CaCO_3}-m_{Ca\left(OH\right)_2}=0,075\cdot100-0,075\cdot40=4,5g\)

- Nếu X chỉ có CaCO3

n CaCO3 = 20/100 = 0,2(mol)

CaCO3 + 2HCl $\to$ CaCl2 + CO2 + H2O

n HCl = 2n CaCO3 = 0,4(mol)

=> mdd HCl = 0,4.36,5/20% = 73(gam)

=> V = 73/1,2 = 60,83(ml)

Nếu X chỉ gồm KHCO3

n KHCO3 = 20/100 = 0,2 mol

KHCO3 + HCl $\to$ KCl + CO2 + H2O

n HCl = n KHCO3 = 0,2 mol

=> mdd HCl = 0,2.36,5/20% = 36,5 gam

=> V = 36,5/1,2 = 30,42(ml)

Vậy : 30,42 < V < 60,83

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