Question

Identify the limiting reagent and the moles of urea, (NH2)2CO(aq), produced given the following balanced chemical reaction and initial amounts: 2NH3(g)+ CO2(g) → (NH2)2CO(aq) + H2O(l) initial NH3(g) (17.03 g/mol): 15.33 g initial CO2(g) (44.01 g/mol): 0.600 moles

Answer 1

Limiting reagent: NH3

moles of Urea = 0.45 moles

Step-by-step explanation:

To do this, all we need to do is to see the balanced reaction which is:

2NH3 + CO2 ----------> (NH2)2CO + H20

Now, the leading numbers in the molecules mean in other words, the theorical moles that are needed to get this reaction.

So according to the reaction, 2 moles of NH3 reacts with 1 mol of CO2. Now, let's see how many moles do we actually have with the initial data:

moles of NH3 = 15.33/17.03 = 0.9 moles

moles of CO2 = 0.6 moles

Now that we have the moles, let's see which is the limiting reagent:

If:

2 NH3 -------> 1 CO2 then:

0.9 -----------> X

Solving for X:

X = 0.9 * 1 / 2 = 0.45 theorical moles of CO2

And we have 0.6 moles of CO2, which mean that we have in excess the CO2, therefore, the limiting reagent is the NH3. Let's prove it:

2NH3 -------> 1 CO2

X ----------> 0.6 moles

X = 0.6 * 2 / 1 = 1.2 theorical moles of NH3

And we have 0.9 moles of NH3. So this is a proof that NH3 is the limiting reagent as we state before.

As the NH3 is the limiting reagent, this means that in the reaction all the moles of NH3 will be consumed in the reaction and then, it will produce moles of urea.

Remanent moles of CO2 = 0.6 - 0.45 = 0.15 moles of CO2

moles of NH3 at the end of reaction: 0

moles of urea = 0.45 moles