Ca (hso3) Add excess naoh ion equation nahco3 Add a small amount of ca (oh)2 ion equation

Ca (hso3) Add excess naoh ion equation nahco3 Add a small amount of ca (oh)2 ion equation

To solve this kind of problem, we can set the quantity of insufficient substance as little as one, and the quantity of other substances is unlimited:
The first problem is that if there is only one Ca (HSO3)2, H+ will react with OH- to form water, Ca2+ and (SO3)2- to form CaSO3↓, the remaining is Na2SO3, and the ion equation is (Ca2+)+2(( HSO3)-)+2OH-=CaSO3 2H2O+(SO3)2-;
The second problem: Let Ca (OH)2 have only one, then H+ will react with OH- to form water, Ca2+ and (CO3)2- to form CaCO3↓, the remaining is NaOH, the ion equation is (HCO3-)+Ca (OH)2=CaCO3↓+H2O+OH-;
Or write the chemical equation and rewrite it as the ion equation.

Why does NaHCO3 react with Ca (oH)2?

This is the reaction of an acid salt with a base.
This reaction is a metathesis reaction, mainly consisting of a small amount and a small amount as one.
When there is a small amount of calcium hydroxide, the stoichiometric number of calcium hydroxide is 1, and each mole of calcium hydroxide contains 2 moles of hydroxide ion. If the hydroxide ion is completely neutralized,2 moles of hydrogen ion is required. Therefore, when the stoichiometric number of sodium bicarbonate is added, there is no sodium hydroxide in the product. The chemical equation of the reaction is:
2NaHCO3+Ca (OH)2=CaCO3 Na2CO3+2H2O
When the calcium hydroxide is excessive, the product contains sodium hydroxide, and the chemical equation of the reaction is:
NaHCO3+Ca (OH)2=CaCO3 NaOH+H2O
Hope my answer can be helpful to your study!