Question

3cu(s)+8HNO3=3cu(No3)2+2NO+4H2O(1)
How many nitric oxide, NO are formed after mixing if we have 0.25mol cu with 0.7 mol HNO3

Answer 1

I have rewritten the equation to help understand it. I assume you want to know number of mols NO formed under these conditions.

3Cu(s) + 8HNO3 => 3Cu(NO3)2 + 2NO + 4H2O(l)
mols Cu = 0.25
mols HNO3 = 0.7
mols NO from Cu if exess of HNO3 =
0.25 x (2 mol NO/3 mols Cu) = 0.25 x 2/3 = 0,167

mols NO from HNO3 if Cu in excess =
0.7 x (2 mols NO/8 mols HNO3) = 0.175

In limiting reagent problems the SMALLER number always wins (you can't get more than the smallest) so Cu is limiting reagent and you will form 0.167 mols NO.