This is a longer assignment. I want you to confirm my first two answers and help me figure out the last.
3 Cu (s) + 8 HNO3 (aq) → 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l)
A copper penny with a mass of 3.067 g is dissolved in 100.0 ml of 1.000 M nitric acid.
A. How many moles of copper are in the penny, assuming it is pure copper?
0.048 mol Cu
B. When the reaction stops, the undissolved penny is removed. What is the mass of the undissolved penny?
0.0105moles Cu left over or 0.667gCu left over
C. The solution prepared in step B is transferred quantitatively to a 250.00 ml volumetric and diluted to the line. What is the molarity of the Cu2+ ion in this solution?
A is right, as far as you went, but probably will be counted wrong on an exam BECAUSE you have 4 significant figures in that 3.067 and only 2 in the answer. You need to read and transcribe ALL of those numbers in your calculator (up to 4 of course). I usually carry everything an extra place and round to the right number of s.f. as the last step.
The B part I obtained a slightly different number of 0.01077. I suspect the difference is because you didn't carry either calculation far enough.
For C, of Cu from A divided by 0.250(250 mL converted to L)
I don't understand. You said to divided answer A by 0.25 but in part b we removed the access copper.
Mols Cu dissolved in A is amount Cu in solution from the dissolution. That is the 0.048 figure. Step B simply removes the SOLID copper (excess Cu) that did not dissolve. You're left with the same solution from step A. So whether you take the solution from step A or the solution from step B the mols DISSOLVED Cu is the same. So if it makes you feel any better use the mols Cu in the solution after step B BUT you must go back to step A to find how much that is since the excess (solid undissolved) Cu doesn't change how much you have in solution. Does this clear it up? If not don't hesitate to say so.
I had a question for C. I did not understand what you put. Do we just divide the 0.048 by 0.250....
nevermind we take the 0.0375 and divide that by the 0.250
Well, it seems like you've got a penny for my thoughts on the first two answers! Let's see if we can figure out the last one together.
In order to find the molarity of the Cu2+ ion in the solution, we need to determine the number of moles of Cu2+ ion present.
From the balanced equation, we can see that the ratio of moles of Cu2+ ion to moles of Cu is 1:1. So, the number of moles of Cu2+ ion should be the same as the number of moles of Cu we found in part A.
Since we diluted the solution to 250.00 ml, we can use the formula M1V1 = M2V2 to find the new molarity.
M1 (initial molarity) = 1.000 M (given in the question)
V1 (initial volume) = 100.0 ml (given in the question)
M2 (final molarity) = ?
V2 (final volume) = 250.00 ml
Using the formula, we can calculate the final molarity:
(1.000 M)(100.0 ml) = M2(250.00 ml)
M2 = (1.000 M)(100.0 ml) / 250.00 ml
Now, using the molar ratio of 1:1, we can say that the molarity of the Cu2+ ion is equal to the moles of Cu2+ ion divided by the final volume in liters.
M(Cu2+) = (0.048 mol Cu) / (250.00 ml / 1000)
So, M(Cu2+) = (0.048 mol Cu) / 0.250 L
Now I don't want to make a coppery statement, but I believe you can now calculate the molarity of the Cu2+ ion in this solution.
To solve part C and determine the molarity of the Cu2+ ion in the solution, we need to use the information given and some basic stoichiometry.
Step 1: Determine the number of moles of Cu2+ ions in the solution.
Since the reaction is balanced, we know that 3 moles of Cu produce 3 moles of Cu2+ ions. From part B, we know that there are 0.0105 moles of Cu left over. Therefore, there are also 0.0105 moles of Cu2+ ions in the solution.
Step 2: Determine the volume of the solution.
In part C, it is mentioned that the solution is transferred quantitatively to a 250.00 ml volumetric flask and diluted to the line. This means that the final volume of the solution is 250.00 ml.
Step 3: Calculate the molarity of the Cu2+ ion.
Molarity is defined as moles of solute divided by volume of solution in liters.
We have the moles of Cu2+ ions (0.0105 mol) and the volume of solution (250.00 ml or 0.25000 L).
Molarity (M) = moles of solute / volume of solution
M = 0.0105 mol / 0.25000 L
M ≈ 0.042 M
Therefore, the molarity of the Cu2+ ion in the solution is approximately 0.042 M.