Two chemical equations for the reaction of CaHCO32 with a certain amount and a sufficient amount of

The chemical equation has only one Ca(HCO3)2+2HCl==CaCl2+2CO2+2H2O

The ionic equation is also the same.

A small amount of HC3-+H+==CO2+H2O sufficient amount: 2HCO3-+2H+==2CO2+2H2O about 2 as above.

This is just the reaction between bicarbonate and hydrogen ions, but when there are small amounts of hydrogen ions, the carbonic acid in the water will not overflow in the form of carbon dioxide, so it is carbonic acid, but a large number of hydrogen ions will become carbon dioxide.

So the first one is HCO3 +H+=H2CO3. The second is HCO3-+H+==CO2+H2O

Ca(HCO3)2 + NaOH ==CaCO3 + NaHCO3 + H2O (small amount of NaOH).

Ca(HCO3)2 + 2NaOH ==CaCO3 + Na2CO3 + 2H2O (adequate NaOH).

Ionic Equation:

Ca2+ +HCO3- +OH- =CaCO3 + H2O

Ca2+ +2HCO3- +2OH- =CAC3 + CO32- +2H2O (sufficient NaOH).

23) Ca(OH)2 + NaHCO3 ==CaCO3 + NaOH + H2O (a small amount of coarse osmotic car NaHCO3).

Ca(OH)2 + 2NaHCO3 ==CaCO3 + Na2CO3 + 2H2O (adequate NaHCO3).

Ionic Equation:

Ca2++OH- +HCO3- = CaCO3 + H2O (a small amount of NaHCO3).

Ca2+ +2OH- +2HCO3- =CAC3 + CO32- +2H2O (shouting sufficient amount of NaHCO3).

Add a small amount of NaOH solution to a sufficient amount of Ca(HCO3)2 solution, and the ion equation of the reaction:

ca2+ +hco3- +oh-=caco3↓+h2o

Ca(HCO3)2 reacts with NaOH.

Overdose: Ca(HCO3)2+2NaOH=CaCO3 +2H2O+Na2CO3

Ion: Ca2+ +2HCO3- +2OH- = CaCO3 + 2H2O+CO32-

The amount of ca(HCO3)2+NaOH=CaCO3 + H2O+NaHCO3

Ion first sensitive: Shenkai Ca2++ HCO3- +OH-=CaCO3 + H2O

Simply read it clearly. A small amount of coefficient 1Here's the equation: Fight Wild.

ca(hco3)2+ca(oh)2=2ca（co3）+2h2o

Ca+HCO3+OH = Ca(CO3)+H2O(Plantae)

ca(hco3)2 + na2co3 ==caco3↓ +2nahco3

ca(hco3)2 + ca(oh)2 = 2caco3↓ +2h2o

naoh + hcl = nacl + h2o

ca(oh)2 + 2hcl = cacl2 + 2h2o

na2co3 + ca(oh)2 = caco3↓ +2naoh

Ca(HCO3)2 + 2HCl ==CaCl2 + 2CO2 Chaos + 2H2O

Add sodium carbonate dropwise to hydrochloric acid: Yanyu Na2CO3 + 2HCl ==2NaCl + CO2 + H2O

Add hydrochloric acid dropwise to sodium carbonate: Na2CO3 + HCl ==NaCl + NaHCO3

nahco3 + hcl ==nacl + co2↑ +h2o

1) Add a sufficient amount of Ca(OH)2 solution to the Ca(HCO3)2 solution.

Ca(HCO3)2 + Ca(OH)2 = 2CaCO3(precipitate) + 2H2O

2) Ionic equations.

2Ca2+ 2HCO3- +2OH- =2CaCO3 (precipitation) slag + 2H2O

Ca(2+)+HCO3-+OH-=CaCO3 submerged chain noisy lake +H2O

PS: If you're like you, there will still be ca(2) leftover, and the shack will continue to reflect the generation of caco3 with co3(2—).

Because pure knowledge of NaOH is too much to do the smuggling, that is, the amount of Ca(HCO3)2 is insufficient.

ca(hco3)2+2naoh=na2co3+2h2o+caco3↓

ca(hco3)2+2naoh===caco3↓+na2co3+2h2o

The chemical reaction equation: Ca(HCO3)2 + Ca(OH)2 == 2CaCO3(precipitation) + 2H2O

Ion equation: 2Ca2+ +2HCO3- +2OH- == 2CaCO3(precipitation) +2H2O

Because they are all Ca2+, the generated must be CaCO3 precipitation, so there is no problem of excess.

But it would be different if CA(OH)2 was changed to NAOH.

A small amount of NaOH: Ca(HCO3)2 + NaOH == NaHCO3 + CaCO3(precipitate) + H2O

NaOH excess: Ca(HCO3)2 + 2NaOH == Na2CO3 + 2CaCO3(precipitate) +2H2O