Balancing Reaction of Formic Acid and Ammonium Hydroxide

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The chemical reaction between 1 mole of ammonia (NH3) and 0.1 mole of formic acid (HCO2H) produces ammonium formate (NH4HCO2) without generating water. The balanced equation is NH3 + HCOOH → HCOONH4. In this reaction, there is a 1:1 mole ratio, and since only 0.1 mole of formic acid is present, 0.9 moles of ammonia remain unreacted. It's important to note that when writing balanced equations, only stoichiometric coefficients should be used, not specific mole quantities.
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I'd like to write the chemical equation for the reaction of 1 mole of ammonium hydroxide with .1 mole of formic acid. I understand that NH4OH is not ammonium hydroxide, and it is more accurate to write NH3, implying NH3 (ammonia) + HCO2H (formic acid) --> NH4HCO2 (ammonium formate) but would there also be water produced from the reaction? Also this reaction has a 1:1 mole ratio for the ammonia and formic acid so would I also add a term on the right for excess ammonia that does not react? Thanks for the help - I haven't had a chemistry course in years.
 
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PCSL said:
implying NH3 (ammonia) + HCO2H (formic acid) --> NH4HCO2 (ammonium formate)

That's a reasonably correct depiction of what is happening. There is no water produced.

would I also add a term on the right for excess ammonia that does not react?

No. Google limiting reagent.
 
Borek said:
No. Google limiting reagent.

Perfect - so I could just write NH3+.1*HCO2H --> .1NH4HCO2 with .9 moles of ammonia in excess? Thanks.
 
Not exactly. You don't use numbers of moles in the balanced reaction - only stoichiometric coefficients. So your reaction, when balanced, should look like

NH3 + HCOOH → HCOONH4

(I am used to a bit different convention for writing carboxylic acids and their salts, nothing wrong with the one you use though).

And then you can describe what is happening in your particular case. Eons ago I was trained to write it below the reaction equation (but formatting such things these days is getting ridiculously time consuming).

And yes, there is 0.9 moles of the ammonia left after the reaction.
 
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