NaHCO3 reacts with clarified lime water What is the single mole of a small amount?

Because the reaction equation is different when the clarified lime water is insufficient and the sodium bicarbonate is insufficient, the equation is Ca2+ +2OH- +2HCO3- ==== CaCO3 +2H2O + CO3(2-)

In this case, the equation is HCO3- +CA2+ +OH- ==== CAC3 +H2O

Therefore, in order to write similar equations more accurately, I found the rule, that is, write the insufficient substance first, and write it strictly according to its ion ratio, for example, calcium hydroxide should write ca2+ +2oh- and not ca2+ +oh- After writing, then see how 2 hydroxides can react with several bicarbonates, and write a few bicarbonates, regardless of whether the excess substances are written according to the chemical ratio. This is called a small amount of 1mol

Hello, this kind of reaction teacher has talked about.

Due to the different dosages of NaHCO3 and Ca(OH)2, the equations written are also different.

So write the coefficient of the small amount as 1 and the other reactant as needed, and then you can write the correct equation

Remember the principle that "whoever has less will set 1" is very useful!

Oh, brother, I am studying chemistry, welcome to ask o(oha!

The upstairs is very good, this is a very important knowledge point, and it has applications in the hydrolysis of salts.

The ionic equation for the reaction of NaHCO3 solution with a small amount of clarified lime water is: 2HCO3-+Ca2++2OH- ==CaCO3 +H2O+CO32-.

You can understand this in this way, a small amount of clarified limestone reacts with it, that is, sodium bicarbonate is excessive, and there must be sodium bicarbonate left after the reaction, and the product must not react with sodium bicarbonate at this time.

The ionic equation for the reaction of NaHCO3 solution with excess clarified lime water is: HCO3-+Ca2++OH- ==CaCO3 +H2O. A sufficient amount of clarified limestone reacts with it, that is, the clarified limestone is excessive, and there must be a surplus of clarified limestone after the reaction, and the product must not react with the clarified limestone at this time.

Therefore, the ionic reaction equation is different.

A small amount of lime water has an excess of hydrogen ions, and the calcium ions react completely.

Excess lime water hydroxide ions are excessive, and hydrogen ions react completely.

Excess NaHCO3 produces CaCO3 and NaOH

There is less NaHCO3, resulting in CaCO3 and Na2CO3

Upstairs, let me explain: due to the excess of lime water, it can ensure that all bicarbonate participates in the reaction.

Every 2mol of HCO3- is ionized with 1mol of Ca(OH)2 and 2molOH2 to form 2mol of CO32 and 2mol of H2O, of which 1mol of CO32 is combined with Ca2+ to form 1mol of CaCO3 precipitate.

2nahco3+ca(oh)2===caco3↓+na2co3+2h2o

The ionic equation is.

2hco3-+ca2++2oh-==caco3↓+co32-+2h2o

Let the mass of Na2CO3 be X, and the mass of NaHCO3 is 100GCaCo3 precipitate requires CO244G, i.e., Element C is required 12GC element conservation: Element C in Na2CO3: X*12 106 Element C in NaHCO3: (

x*12/106 +

x=naHCO3

Let the amount of Na2CO3 be x mol, and NaHCO3 be y molNa2CO3 - CO2; nahco3——co2；Both CO2 and CACO3 are 1:1

106x+84y= 44*(x+y)=44 x=na2CO3 is and NaHCO3 is.

A Reason: The molar mass of Na2CO3 and NaHCO3 solids of equal mass is 106g mol and the molar mass of NaHCO3 is 84g mol, so the amount of NaHCO3 is large and the amount of CO2 emitted is more.

The team will answer the following questions for you: a

Because the ionic reaction of the metathesis reaction type is mainly based on a small amount, and the small amount is 1, that is, the stoichiometric number of a small amount is 1, the ionic equation of this reaction should be: HCO3 - CA2+ +OH - = CaCO3 +H2O

The chemical equation is: NaHCO3 + Ca(OH)2= CaCO3 + NaOH + H2O

I don't know if I've learned it before, first of all, the left side should be a CA ion, and the right side of the carbonate should be negative 2 valence.

The amount of the substance that is calcium carbonate is.

100g mol, so the carbon dioxide produced by heating sodium bicarbonate is, then:

2nahco3

na2co3+h2o+co2↑

Therefore, the mass of NaHCO3 in the original mixed solid is , and it can be seen from the equation that the sodium carbonate generated by the decomposition of sodium bicarbonate by heating is that, and the amount of carbon dioxide generated by the reaction of the remaining solid matter after heating with a sufficient amount of hydrochloric acid is =, then:

Na2CO3+2HCl 2NaCl+H2O+CO2 Therefore, the amount of Na2CO3 in the original mixed solid is, so the mass of Na2CO3 in the original mixed solid is, Answer: The mass of NaHCO3 in the original mixture is, and the mass of Na2CO3 is.