# Solve Net Ionic Equation of NaOH + (NH4)2CO3 Reaction

• leebongemail
In summary: In summary, when a solution of sodium hydroxide is added to a solution of ammonium carbonate, and the solution is heated, ammonia gas, (NH3) is released.
leebongemail
Problem: When a solution of sodium hydroxide is added to a solution of ammonium carbonate, and the solution is heated, ammonia gas, (NH3) is released. Write a net ionic equation for this reaction. Hint: both NaOH and (NH4)2CO3 exist as dissociated ions in aqueous solution.

Can someone help me through the steps to find the net ionic equation?
I need help to begin writing the balanced equation for this system.

First, you need to write a balanced molecular equation.
Then, write an ionic equation that includes all the ions in the reaction.
Third, cross out the spectator ions on both sides of your equation and you get the net ionic equation.
Remember to check solubility rules.

Hint: It is a double displacement reaction.

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Hm, well for the first step i got

NaOH+(NH4)2CO3 <---->NaCO3+NH3+H20

is that correct?

The sodium ion has a 1+ charge whereas the carbonate ion has a 2- charge, so your equation is wrong. (It's Na2CO3)

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so the equation is NaOH+(NH4)2CO3 <---> Na2CO3+NH3+H2O?

Yes, but you have to balance the equation.

2NaOH+(NH4)2CO3 <---> 2Na2CO3+2NH3+2H2O

is this balanced equation correct?

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Hint: Count the number of $$Na^+$$ that you have in both members.

If you have difficulties balancing chemical equations, you may want to see-
Wikipedia said:
Linear system balancing

In reactions involving many compounds, balancing may get harder, we can then try to balance equation using algebraic method, based on solving set of linear equations:

1. Assign variables to each coefficient:

* a K4Fe(CN)6 + b H2SO4 + c H2O → d K2SO4 + e FeSO4 + f (NH4)2SO4 + g CO

2. We must have the same quantities of each atom in each side of the equation. So, for each element, count its atoms and equal both sides:

* K: 4a = 2d
* Fe: 1a = 1e
* C: 6a = g
* N: 6a = 2f
* H: 2b+2c = 8f
* S: b = d+e+f
* O: 4b+c = 4d+4e+4f+g

3. Solving the system (usually direct substitution is the best way)

* d=2a
* e=a
* g=6a
* f=3a
* b=6a
* c=6a

which means that we have all coefficients depending on a parameter a, just choose a=1 (a number that will make all of them small whole numbers) and you'll have:

* a=1 b=6 c=6 d=2 e=1 f=3 g=6

4. And the balanced equation at last:

* K4Fe(CN)6 + 6 H2SO4 + 6 H2O → 2 K2SO4 + FeSO4 + 3 (NH4)2SO4 + 6 CO
I always use this method when I can't figure out the stoichiometric coefficients by trial-error.

http://en.wikipedia.org/wiki/Chemical_equations
http://www.studyworksonline.com/cda/content/article/0,,EXP1315_NAV2-100_SAR1316,00.shtml

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where did K come from?

leebongemail said:
Problem: When a solution of sodium hydroxide is added to a solution of ammonium carbonate, and the solution is heated, ammonia gas, (NH3) is released. Write a net ionic equation for this reaction. Hint: both NaOH and (NH4)2CO3 exist as dissociated ions in aqueous solution.

Can someone help me through the steps to find the net ionic equation?
I need help to begin writing the balanced equation for this system.

First you need to know what ions your solution contains. Ammoniumcarbonate solution contains:

$$NH_{4} ^{+}$$ and $$CO_{3} ^{2-}$$

Sodium hydroxide solution contains:

$$Na^{+}$$ and $$OH^{-}$$

We already know that $$NH_{3}(g)$$ will be formed and so you know that the ammonium ions will react with the hydroxide ions. The sodium carobonate that will be formed is soluble in water and thus don't need to be taken up in the ionic equation. Thus the net ionic reaction occurring here is:

$$NH_{4} ^{+} + OH^{-} \rightarrow NH_{3} + H_{2}O$$

remember that you added 2 ion conataining solution with each other, thus the ions already exist. When solid ammonium carbonate is added to a sodium hydroxide solution, you refer to $$(NH_{4})_{2}CO_{3}$$ in the equation, because it needs to be dissolved first.

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is there a website that has the common or many ions for an element? I'm taking a summer introduction to chem class at this community college, and it goes by really fast. does someone recommend a site? i can sure use it. I have a test tomorrow on 'Acids and Bases' and 'Reaction Rates and Chemical Equilibrium'. I'm sure to have questions later in the day!

Here's a site that has a list of common ions: http://www.launc.tased.edu.au/online/sciences/PhysSci/pschem/ion/ion_tabl.htm

Here's a website that I've found to be very useful for my summer chemistry class: http://wine1.sb.fsu.edu/chm1045/chm1045.htm

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## What is a net ionic equation?

A net ionic equation is a chemical equation that only shows the participating ions and excludes any spectator ions. Spectator ions are those that do not participate in the reaction and remain in their original form.

## How do I determine the net ionic equation of a reaction?

To determine the net ionic equation, first write out the balanced molecular equation for the reaction. Then, use solubility rules to determine which ions are soluble and which ones are insoluble. The insoluble ions will be the ones that form the solid or precipitate in the net ionic equation. Finally, write out the net ionic equation by removing any spectator ions from the balanced molecular equation.

## What is the balanced molecular equation for the reaction between NaOH and (NH4)2CO3?

The balanced molecular equation for the reaction between NaOH and (NH4)2CO3 is: 2NaOH + (NH4)2CO3 → 2NH4OH + Na2CO3

## Which ions are soluble in the reaction between NaOH and (NH4)2CO3?

In this reaction, the ions that are soluble are Na+ and OH-. These ions remain in solution and do not form a solid or precipitate. The ions that are insoluble are NH4+ and CO3^2-, which form the precipitate NH4OH.

## What is the net ionic equation for the reaction between NaOH and (NH4)2CO3?

The net ionic equation for the reaction between NaOH and (NH4)2CO3 is: 2NH4+ + 2OH- → 2NH4OH

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