Question from test 1 [ENDORSED]

[Zuri Smith 1A]

I don't know if we're allowed to ask questions from test 1, and so I sincerely apologize if I am breaking a rule by posting this question. I was wondering how to do the following, which involves finding the limiting reaction and then determining how much unreacted fuel is left over. The equation is 2H2 (g) + O2 (g) --> 2H2O (g). They give us:
"A rocket carries 100.00 g of H2 and 100.0 g of O2 What is the limiting reagent? And how much unreacted fuel is left over?"
I understand how to find the limiting reagent, which I determined to be Oxygen in part A. However, I do not understand how to do the second part.

[Odalys Cuevas 1C]

I think that to do the second part you would take the mol you got from part a that allowed you to determine that 02 was the limiting reagent and turn that to grams of H2. Then you would subtract the grams given by the grams you found to get what is left over of the unreacted one. Thats how we did it in peer learning, so hopefully it helps.

[Joshua Yang 1H]

I'm sure we can discuss this since we already discussed test 1 in our discussion section

So the calculation would be...

Molar Mass of H2 : 2.01588 g/mol
Molar Mass of O2 : 31.9988 g/mol

Since O2 is the limiting reactant


100.00g H2 / (2 x 2.01588 g/mol) - 100.00g O2 / (31.9988 g/mol ) = mol of H2 leftover


then to get the mass of H2 leftover, simply

mol of H2 leftover x 2.01588 g/mol = answer

which will give you about 43.70013876 round to 5 sig figs and you get 43.700g H2

[Zuri Smith 1A]

Okay thank you! And just to reconfirm, the limiting reagent is basically just the reactant that has the smallest # of moles (and thus runs out the quickest)?

[Jazmin 1H]

Okay thank you! And just to reconfirm, the limiting reagent is basically just the reactant that has the smallest # of moles (and thus runs out the quickest)?

I would say that you are correct about the limiting reactant.