Asked by michelle
Hydrogen Chloride and Sodium Carbonate react to produce Hydrogen Carbonate and Chloride Sodium how do i write that in net ionic form?
Answers
Answered by
DrBob222
Step 1. Write the balanced molecular equation.
I assume this is in aqueous solution.
2HCl(aq) + Na2CO3(aq) ==> 2NaCl(aq) + H2CO3(aq)
Step 2. Convert these to ions unless the material is
a. a gas,
b. a solid (insoluble ppt),
c. a weak electrolyte.
If any of those, write it as the molecule and not as the ion.
For example, HCl(aq) would be written as H^+(aq) + Cl^-(aq)... etc.
Step 3. Cancel ions common to both sides. What remains is the net ionic equation.
Note: I don't know how your teacher wants H2CO3 handled. Personally, I would show it as CO2(g) + H2O(l). If you DO NOT decompose the H2CO3, then H2CO3 is written as the molecule in the net ionic equqtion. If you do, then both CO2 and H2O are written as the molecule.
Be sure you know how to do these and post a follow up question if you don't understand. The net ionic equation is
2H^+(aq) + CO3^=(aq) ==> CO2(g) + H2O(l)
I assume this is in aqueous solution.
2HCl(aq) + Na2CO3(aq) ==> 2NaCl(aq) + H2CO3(aq)
Step 2. Convert these to ions unless the material is
a. a gas,
b. a solid (insoluble ppt),
c. a weak electrolyte.
If any of those, write it as the molecule and not as the ion.
For example, HCl(aq) would be written as H^+(aq) + Cl^-(aq)... etc.
Step 3. Cancel ions common to both sides. What remains is the net ionic equation.
Note: I don't know how your teacher wants H2CO3 handled. Personally, I would show it as CO2(g) + H2O(l). If you DO NOT decompose the H2CO3, then H2CO3 is written as the molecule in the net ionic equqtion. If you do, then both CO2 and H2O are written as the molecule.
Be sure you know how to do these and post a follow up question if you don't understand. The net ionic equation is
2H^+(aq) + CO3^=(aq) ==> CO2(g) + H2O(l)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.