using your initial mass of copper, calculate the minimum volume of concentrated 12M nitric acid needed to completely react with your copper to form copper(II) nitrate? Original mass of copper was 0.40g.

Answer 1

I will assume you are using concentrated HNO3. With concd HNO3 the reduction product is NO2; with dilute HNO3 the reduced product is NO.
Cu + 4HNO3 ==> Cu(NO3)2 + 2H2O + 2NO2

mols Cu = grams/molar mass = approx 0.0063 but you need to do it and the calculations that follow more accurately because I've estimated the numbers.

Convert mols Cu to mols HNO3. That's 0.0063 x (4 mol HNO3/1 mol Cu) = approx 0.025

Then M HNO3 = mols HNO3/L HNO3. You know M HNO3 from the problem and mols from the above calculation; solve for L: HNO3 and convert to mL.