Find the mininum ammount of 5.0 M HNO3 needed to dissolve 2.72 g of Cu (s). I have the balanced equation here:

2NO3-(aq)+8H+(aq)+3Cu(s)-->2NO(g)+4H2)(l)+3Cu2+(aq)

Obviously it is a 2 moles NO3- to 3 moles of Cu (s) ratio. I converted grams Cu to moles (0.04280)), then multiplied by 2/3 (ratio of NO3- to Cu) which gave me moles of NO3- (0.02853). I then solved for the volume required by dividing those moles by 5.0 M, however when I submit the answer I is not correct. Can anyone see my error? The answer I am getting is 5.7 mL.

You have balanced the redox part.

The molecular equation is
8HNO3 + 3Cu ==> etc
Try that and get back either way, please.

Yes you were correct, it is an 8/3 ratio. Thanks Dr. Bob!!

To find the minimum amount of 5.0 M HNO3 needed to dissolve 2.72 g of Cu(s), you correctly identified the balanced equation:

2NO3-(aq) + 8H+(aq) + 3Cu(s) --> 2NO(g) + 4H2)(l) + 3Cu2+(aq)

Next, you converted the mass of Cu to moles: 2.72 g Cu * (1 mol Cu / molar mass of Cu) = 0.04280 mol Cu.

Then, you correctly recognized the stoichiometry of the reaction: 2 moles of NO3- are required to dissolve 3 moles of Cu. So, you multiplied the moles of Cu by the ratio 2/3 to find the moles of NO3-: 0.04280 mol Cu * (2 mol NO3- / 3 mol Cu) = 0.02853 mol NO3-.

Finally, to find the volume of 5.0 M HNO3 required, you divided the moles of NO3- by the molarity of HNO3: 0.02853 mol NO3- / 5.0 mol/L = 0.005706 L, which is equivalent to 5.706 mL.

Based on your calculations, the answer you obtained is indeed 5.7 mL. However, it's important to consider significant figures. Since the original mass of Cu is given with four significant figures (2.72 g), it's advisable to use all four significant figures in the calculations to maintain accuracy.

If your answer is not being accepted as correct, double-check whether the given information requires a different number of significant figures or if there might be a small rounding error in your calculations.