Very confused, how is Copper(II) Oxide in excess??

So basically for CuO moles= 5.60/79.5 = 0.07044025157 mol
and HNO3 moles - 0.5 x 2.50 = 1.25 mol

so what I did was to find the limiting reagent!
so CuO : HNO3
1 : 2
so to find the moles of HNO3, I multiplied 0.07044025157 x 2 = 0.140880.. so basically HNO3 must be the one in excess, am I not right?!

Now, to find the moles of CuO, 1.25/2 = 0.625 , so CuO is not the one in excess but the limiting reagent!!!

So could anyone please tell me why does question say that CuO is in excess when it's clearly HNO3 the one that's in excess? :s-smilie:

(edited 8 years ago)

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Reply 1

Nitrogen

1st of all you got the moles of HNO3 wrong. To convert cm3 to dm3 you divide by 1000 not 100. The mol of HNO3 that reacted is actually 0.125. You know that the ratio between the reactants is 1:2 which means you only need half as much CuO compared HNO3. 0.125/2 = mol of CuO needed. Compare the value with mol of CuO used and you can see it is in excess.