Chemistry
posted by J.J on .
For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
4Cr(s)+3O2(g)→2Cr2O3(s)
1.) 1 molCr; 1 molO2
2.) 4 molCr; 2.5 molO2
3.) 12 molCr; 10 molO2
4.) 14.8 molCr; 10.3 molO2
Express your answer as a chemical formula.
I'm stuck on how you write it as a chemical formula, pleaseee help! Thanks!

I've never done this but I would guess it would be Cr4O3; i.e., for every 4 mols Cr you use 3 mol O2 for #1 with Cr as the limiting reagent.
For #2, O2 is the LR.
2.5 O2 x 4 mol Cr/3 mol O2 = 3.33 so the formula is Cr3.33O2.5. Convert to whole numbers and that is Cr10O7.5 then double that to get rid of 1/2 which makes it Cr20O15. That is divisible by 5 for Cr4O3. ta dah!
I don't know if this is what you want or not.
#3 doesn't work out to be the same as 1 and 2.