$$\ce{2NH3 + CO2 -> (NH2)2CO + H2O}$$

If we have $637.2\ \mathrm{g}$ of $\ce{NH3}$ and $1142.0\ \mathrm{g}$ of $\ce{CO2}$ how many grams of $\ce{(NH2)2CO}$ will we get?

This is the explanation I was given:

First, convert reactives to moles:

  • $637.2\ \mathrm{g}$ of $\ce{NH3}$ are $37.5$ moles.
  • $1142.0\ \mathrm{g}$ of $\ce{CO2}$ are $26$ moles.

Second, define the stoichiometric proportion between reactives and products...

  • From $2$ moles of $\ce{NH3}$ we get $1$ mol of $\ce{(NH2)2CO}$
  • From $1$ mol of $\ce{CO2}$ we get $1$ mol of $\ce{(NH2)2CO}$

Third, calculate the number of product moles we would get if each reactive was completely consumed:

  • Consuming all the $\ce{NH3}$ we would get $18.75$ moles of $\ce{(NH2)2CO}$
  • Consuming all the $\ce{CO2}$ we would get $26$ moles of $\ce{(NH2)2CO}$

The limiting reagent is $\ce{NH3}$, so the most we will get of $\ce{(NH2)2CO}$ is $18.75$ moles. Now convert that to grams, we get $1125.0\ \mathrm{g}$.

  • Why are we ignoring the existence of $\ce{H2O}$? Does it not matter?
  • IF the formula is unbalanced, do I have to balance it first before making this calculation? What if I can't balance it?

3 Answers 3

  1. Here, $\ce{H2O}$ is a product and you are assuming that the reaction goes to completion, so it does not affect the yield of the reaction in any way. Maybe a suitable analogy is to think of the reaction system as a car - you give the car a certain amount of petrol (say 2 gallons) and it converts it to a certain number of miles (say 70 miles), as well as a lot of exhaust gases. Now the amount of exhaust gases that your car is putting out doesn't affect the fact that if you put in 2 gallons of petrol, you get 70 miles out of it. If you were to treat the reaction as a reversible one, where you have an equilibrium between reactants and products, then in general, the amount of products present would also matter. But don't worry about this for now, I'm sure you will come across this in your study of chemistry at some point in time.

  2. Good question - if the reaction is not balanced, then yes, you do have to balance it by finding the appropriate stoichiometric coefficients. If you don't know exactly how many miles per gallon (35 in this case) your car consumes, then you won't be able to work out how many miles you can drive if you feed your car 2 gallons of petrol. Not all reactions can be balanced, but if a reaction can't be balanced in one way or another, it can't happen in real life. For example, you can't balance the reaction $\ce{Pb -> Au}$, but as we all know, alchemy was (mostly) a load of rubbish. For this reason, any reaction that you come across in your study of chemistry, as well as in real life, can be balanced.


The product $\ce{H_2O}$ does not matter to the question. You have some reactants and the reaction proceeds until one is used up. As you are asked about the mass of one product, you don't care about how much of the other product there is.

Yes, you need to balance the reaction. You need to figure out which reactant is the limiting one. Your reaction is balanced, so you use two molecules of ammonia for each of carbon dioxide. In the answer, under Third, they assume first that $\ce{NH_3}$ is the limiting factor and that there is plenty of $\ce{CO_2}$, finding you can produce $18,75$ moles. Then they assume there is plenty of $\ce{NH_3}$ and you are limited by $\ce{CO_2}$, finding you can produce $26$ moles. The first assumption is correct, so that is how much you can make.


Did anyone explain why you convert to moles? It is because moles is a certain quantity of molecules of and to balance the equation you need to work in molecules or molecule multiples.

So if it goes to completion you will use up all of the $\ce{NH3}$ as you say, but be left with 7.25 moles of $\ce{CO2}$, 18.75 moles of $\ce{(NH2)2CO2}$ and 18.75 moles of water. By the way, the original equation is already stoichiometricly balanced

6 $\ce{H}$ on both sides, 2 $\ce{O}$, and 2 $\ce{N}$ - It's just like algebra.

But if it wasn't you would have to balance it before calculating the rest and of course we are simplifying by assuming the reaction goes to completion.


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