Question

I really need help.
Volume of HCl solution 100.5 mL HCl
Volume of NaOH solution 100.0 mL NaOH
Initial temperature in calorimeter 25.2 °C
Final temperature in calorimeter 28.2 °C
3. Determine the number of moles of NaOH.

4. Determine enthalpy per mole of NaOH. Show all of your work.

Answer 1

Are the two solutions mixed? Is any other information provided?

Answer 2

this is the only info provided and its mixed

Answer 3

What about questions 1 and 2? Are there concentrations for the two solutions?

Answer 4

This cannot be solved without having a way to find out how much HCl or NaOH is present.

Answer 5

1 and 2 are writing a balanced equation and determining the enthalpy change.

Answer 6

Im on part II

Answer 7

Aaaaah see it says both solutions are 0.5 M. Now we're in business haha. I think you'll be able to get moles of NaOH with this information, then enthalpy per mole. By the way, for question 2, the total enthalpy change is -2.514 kJ (for this particular reaction using the volumes of NaOH and HCl you used). This is also what the question asks for. You can't say what it is per mole anyways because you haven't found moles of anything yet.

Answer 8

Did you get the answer?

Answer 9

no i did not, and thanks for revising answer 2

Answer 10

is it 50 mol?

Answer 11

It's 0.05 mol. Remember, C = n/V.

Answer 12

how do i find enthalpy it says its the heat lost or gained so it would be 3?

Answer 13

They want to know the enthalpy per mole NaOH. You have the total enthalpy from question 2 and now you have moles of NaOH.

-2.514 kJ/0.05 mol = -50.28 kJ/mol

Answer 14

thank you for the help i appreciate it greatly, but can you explain to me the conclusion portion?

Answer 15

1. Write out a balanced equation for the reaction you investigated in Part II, including phase symbols.
NaOH(s) + HCl(aq) = NaCl(aq) + H2O(l)
2. Determine the enthalpy change of this reaction, per mole of the balanced equation. Show all of your work.
200 g * 4.18 J (C*g) * (28.2 C – 25.2 C) = 2,508 J = 2.508 kJ
-2,508 * 1kJ = -2.508 kJ
1000 J
100.0 mL of HCl * 1 L HCl solution * 0.50 mol HCl solute = 0.05 mol HCl
1000 mL HCl solution 1 L HCl solution
-2.508 kJ = -50.16 kJ per mole
0.05 mol HCl