Oh sorry I didn't post my work.
Well first of all we are given that the chemical equation ends up like this :
3H2 +O2 ---> 2H2O + H2
We are given the mass of H2 initially is 4.52g and the mass of O2 initially is 24.8g. When I use mole ratios I got the correct ending result of H20 to be 27.12g. When I converted both H2 and O2 to H2O, the mass of H2O produced by H2 was 27.12g and the mass of H2O produced by O2 was 27.9g. This means that the O2 was the excess reagent.
So when I entered those parts of the question in, I got them right, but when it asked for the amount of excess reagent in grams remaining, I can't seem to find the appropriate answer. I know that when doing excess reagent calculation, you take the amount of product that was produce, in this case 27.12g H2O, and then convert it back to the reagent in question(O2). So when I set up a mole ratio it ends up:
(27.12gH2O) x (1 mol H2O)/(18.016 g H2O) x (1 mol O2)/(2 mol H2O) x (32 g O2)/(1 mol O2) = __ g O2 used
Thus 24.8g O2 - __g O2 used = mass of remaining excess.
When I did this I got the answer to be .7147g of excess O2, which works out mathematically if trying to convert it back to grams of H2O. I messed around with the 27.12 and changed it to more exact numbers so I also got .880g O2 and .693g O2.
Is this work the work you wanted me to show?