3Cu(s) + 8HNO3(aq) → 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l)

Question

If 15 g of copper metal was added to an aqueous solution containing 6.0 moles of HNO3, how many moles of NO(g) would be produced, assuming a 59.8 % yield.

Answer 1

YOu need 8/3 moles of nitric acid for each mole of copper
so how many moles of copper?
15g/63.5=.236
so we need 8/3 * .236 of nitric acid, and we have....much more than that, so the reaction is controlled by the mass of Copper.
Now, finally, moles of NO:
we get 2/3 mole of NO for each mole of copper, so with the yield we should get
MolesNO=.59(2/3)(.236)

Answer 2

I dont understand