What volume of dry NO at 25°C and 1.00 atm can be prepared by reacting 6.35 g of Cu with 25.0 mL of 6.0 M nitric acid, HNO3?

Question

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

Answer 1

This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants.
mols Cu = grams/atomic mass = ?
mols HNO3 = M x L = ?

Using the coefficients in the balanced equation, convert mols Cu to mols NO.
Do the same and convert mols HNO to mols NO.
You likely will obtain different values for mols NO which means one of the values is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that is the LR.

Using the smaller value for mols NO, to volume using PV = nRT