Question

Hey Matt can you help me with an issue I’m having for a question?

Here is the question:

If 42.5 g N2 react with 10.1 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?

The unbalanced equation is: N2 + H2 → NH3

I balanced the equation to N2 +3 H2 → 2 NH3

I also changed the given grams to moles so I could compare them

(42.5 g N2) / (28.01344 g N2/mol) = 1.5171 mol N2
(10.1 g H2) / (2.015894 g H2/ mol) = 5.0102 mol H2

But now I am stuck!

Answer 1

Hey, so this is where limiting reactants get a little bit harder!! ☺

So next you need to compare the moles but make a note * to remember to multiply by the coefficient of the balanced equation!!! If you forget to do this then it will definitely mess u up!!

1.5171 moles of N2 would react completely with 1.5171 * (3/1) ← this is the ratio we got from the balanced equation

This = 4.5513 moles H2

So now tell me which one is the limiting reactant!

Answer 2

Ok so there is still more H2 present than that, so H2 would be in excess and N2 is the limiting reactant?!

Answer 3

Exactly! You are getting the hang of this now!

Answer 4

So then next from the balanced equation find the moles of the limiting reactant to the moles produced and multiply by the MW of their product Simple divison

Answer 5

(1.5171 mol N2) * (2 mol NH3 / 1 mol N2) ← ratio * (17.03056 g NH3/mol)= 51.7 g?

Answer 6

Perfect. That’s exactly right. And finally you need to subtract the moles of the limiting reactant from moles of the excess reactant and finally multiply by the excess reactant’s MW.

Answer 7

So (( 5.0102 mol H2 which is the initial) – (4.5513 mol H2 which is reacted on)) * (2.015894 g/mol) = .925g H2 left over?

Answer 8

Yay! You got it! You should write it out in steps format so you see it clearly when you submit it!

Answer 9

1)
N2 + 3 H2 → 2 NH3

2)

(42.5 g N2) / (28.01344 g N2/mol) = 1.5171 mol N2
(10.1 g H2) / (2.015894 g H2/mol) = 5.0102 mol H2

3)

1.5171 moles of N2 would react completely with 1.5171 x (3/1) = 4.5513 moles of H2, but there is more H2 present than that, so H2 is in excess and N2 is the limiting reactant.

4)

(1.5171 mol N2) x (2 mol NH3 / 1 mol N2) x (17.03056 g NH3/mol) = 51.7 g NH3

5)

((5.0102 mol H2 initially) - (4.5513 mol H2 reacted)) x (2.015894 g H2/mol) =
0.925 g H2 left over.

Answer 10

You think that shows all the work and will be enough to be acceptable?

Answer 11

Absolutely!