Question

When solving problems involving stoichiometric coefficients, the first step is to make sure you have a balanced chemical equation. Then, you determine the limiting reagent by using the coefficients from the balanced equation. You can keep track of the amounts of all reactant and products before and after a reaction using an ICF table (as shown in the Simulation). Completing the ICF table will also allow you to determine the limiting reagent, and the amount of product formed is based on assuming that the reaction runs to completion with 100%% yield. Parts A and C explore these steps in more detail. Let us consider another reaction.

Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 2.43 mol NH3 and 4.77 mol O2 react.

Express the amounts in moles to two decimal places separated by a comma.

Answer 1

Let's solve this question

We have the required steps given in the problem itself

First step here is to find the balanced reaction

Balances reaction

= 4NH₃ + 5 O₂ ---> 4NO + 6 H₂O

As we can see left side reactant atoms are equal to right side product atoms indicating it is a balanced reaction.

Now we have 2.43 moles Of NH3

And 4.77 moles of O2

And according to balanced equation NH3 and O2 can react in 4:5 ratio only

So

NH₃:O₂

4:5

2.43: ?

So

? = 2.43× (5/4)

? = 3.03 moles

So as we can see not all O2 is used and the reason for it is limited amount of NH₃

So ammonia is limiting reagent

Now the amount of NO and H₂O formed will be similarly calculated as

According to stoichiometry of balanced reaction

NO = (2.43/4) × 4

= 2.43 moles

And

H₂O = (2.43/4) ×6

= 3.64

So the answer is

2.43, 3.64 moles

I hope this helps. If you have any query or want more detailed explanation, feel free to ask in the comments section below.