Problem Description:

Question

You are a workstudy for the chemistry department. Your supervisor has just asked you to prepare 500mL of 3M HCl for tomorrow's undergraduate experiment. In the stockroom explorer, you will find a cabinet called "Stock Solutions". Open this cabinet to find a 2.0L bottle labeled "11.6M HCl". The concentration of the HCl is 11.6M. Please prepare a flask containing 500 ml of a 3 M (+/- 0.005M) solution and relabel it with its precise molarity. Note that you must use realistic transfer mode, a buret, and a volumetric flask for this problem.
As a reminder, to calculate the volume needed to make a solution of a given molarity, you may use the following formula:
C1V1 = C2V2

1. Calculations: (3 marks)
C1V1 = C2V2
(11.6M)(V1) = (3M)(500ml)
11.6M V1 = 1500(M x ml)
V1 = 129.31ml

Volume of HCl used: 129.31mL

2. How do you calculate your percent error?

% error = (experimental concentration – 3.0 M) / 3.0 M * 100, in absolute value.

Answer 1

I don't see a question.
I don't see anything wrong with what you've done except you didn't explin how to prepare the solution.
You want to add the 129.31 mL of 11.6 molar solution to the 500 mL volumetric flask, add DI water to the mark on the flask, mix, stopper, label.
What you have labeled as absolute error actually is percent error since you have multiplied by 100.

Answer 2

You need1.5 mols of HCl

You have 11.6 mols per liter
1.5 /11.6 = .1293 liters of the strong stuff (agree with you)
add that to .5 - .1293 = .3707 liters of neutral water
and you have 0.5 Liters = 500 mL of 3M HCl

now you need to know that strong bottle with the 11.6 M percent error to get the error in your result I think but anyway what is your measurement error for the .1293 liters?
say for example it was .01 liters (10 mL)
then you would end up with .1393 liters of strong which would be 11.6 liters*.1393 mols/liter = 1.616 mols instead of 1.5 mols
that is 100 (1.616 - 1.5) /1.5 = 7.7 % error