\(a/n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,2mol\\ m_{CaCO_3}=0,2.100=20g\\ b/BTNT\left(O\right):2n_{CO_2}=n_{H_2O}\\ \Rightarrow n_{H_2O}=0,2.2=0,4mol\\ BTNT\left(H\right):2n_{H_2O}=n_{HCl}\\ n_{HCl}=0,4.2=0,8mol\\ V_{HCl}=\dfrac{0,8}{0,4}=2l\)

 Viết phương trình hoá học của XCO3 và Y2(CO3)3 với dung dịch HCl và rút ra nhận xét :

nCO2=nH2O;

naxit=2nCO2

Áp dụng định luật bảo toàn khối lượng, ta có :

m2 muối cacbonat + maxit  =  m2 muối clorua + mCO2 + mH20

m2 muối clorua = 10 + (0,03 x 2 x 36,5) – (0,03 x 44) – (0,03 x 18) = 10,33 (gam)

PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\) (1)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\) (2)

\(Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\) (3) 

\(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\) (4)

Ta có: \(m_{HCl}=10.95.20\%=2,19\left(g\right)\Rightarrow n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)

\(n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

Theo PT: \(n_{HCl\left(1\right)+\left(2\right)+\left(4\right)}=2n_{CO_2}=0,02\left(mol\right)\)

\(n_{H_2O\left(1\right)+\left(2\right)+\left(4\right)}=n_{CO_2}=0,01\left(mol\right)\)

\(\Rightarrow n_{HCl\left(3\right)}=0,06-0,02=0,04\left(mol\right)=n_{H_2O\left(3\right)}\)

⇒ n_H2O = 0,01 + 0,04 = 0,05 (mol)

Theo ĐLBT KL, có: m_hh + m_HCl = m _muối + m_CO2 + m_H2O

⇒ m = m _muối = 2,24 + 2,19 - 0,01.44 - 0,05.18 = 3,09 (g)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

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

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

nCO2= 0,15(mol)

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

m(đá vôi)= mCaCO3= 0,15.100=15(g)

a, CO₂ + Ca(OH)₂ \(\rightarrow\) CaCO₃ + H₂O

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

CO₂ + Ca(OH)₂ \(\rightarrow\) CaCO₃ + H₂O

0,15                        0,15                        ( mol )

\(\Rightarrow m_{CaCO_3}=0,15.100=15g\)

\(PTHH:CO_2+Ca\left(OH\right)_2 \rightarrow CaCO_3 +H_2 O\\ n_{CO_2}=0,15\left(mol\right)\\ m_{đá.vôi}=m_{CaCO_3}=0,15.100=15\left(g\right)\). 

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)

Giả sử có 100g dd CH3COOH 15%

Ta có:

nCH3COOH=\(\dfrac{100.15\%}{60}\)=0,25(mol)

PTHH:Ca(OH)2+2CH3COOH→(CH3COO)2Ca+2H2O

⇒nCa(OH)2=n(CH3COO)2Ca=0,125(mol)

⇒mdd(Ca(OH)2)=\(\dfrac{0,125.74}{X\%}\)=9,25

⇒mdd(spu)=\(\dfrac{9,25}{X\%+100}\)⇔C%=9,875%

\(\dfrac{0,125.158.100}{\dfrac{9,25}{X\%+100}}\)=9,875⇔x=9,25

Na2CO3+2HCl---.>2NaCl2+H2O+CO2

x--------------------------------------x(mol)

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

y---------------------------------------y(mol)

n CO2=6,72/22,4=0,3(mol)

Theo bài ra ta có hpt

\(\left\{{}\begin{matrix}106x+100y=30,6\\x+y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

m Na2CO3=0,1.106=10,6(g)

m CaCO3=0,2.100=20(g)

\(n_{CO2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi số mol Na2CO3 và CaCO3 là a và b

\(\left\{{}\begin{matrix}106a+100b=30,6\\a+b=0,3\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\rightarrow m_{Na2CO3}=0,1.106=10,6\left(g\right)\)

\(\rightarrow m_{CaCO3}=0,2.100=20\left(g\right)\)

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

b) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow m_{CaCO_3}=0,15.100=15\left(g\right)\)

nCO2= 0,15(mol)

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

m(đá vôi)= mCaCO3= 0,15.100=15(g)

nhớ like nha

Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

______0,2_____0,2_________0,2 (mol)

b, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4M\)

c, \(m_{CaCO_3}=0,2.100=20\left(g\right)\)

Bạn tham khảo nhé!