# Calculate Moles of NaOH & Heat: Chemistry Experiment

• m0286

#### m0286

Hello again folks.
Im working on my chemistry work, i just completed an experiment.. Heat of Reaction. Now I am working on some questions for it... now I am stuck
They are asking me, The amount of NaOH moles... I am not sure how to calculate this... throughout the experiment i was mixing 5.5g of NaOH to 200mL of water and to begin with I made a solution 4.00ml 1.0mol/L NaOH solution. Hmm well i have three tables to fill out for 3 different reactions Ill show just the results for the first one, and if someone could help me with this one I am sure i could figure out the other two.

Initial temperature of water: 25 C
Final Temperature of water: 30 C
Mass of NaOH: 5.5g
Volume of water 200mL
Temperature change: 6 C
Mass of solution: 200g ( since i had 1.0mol/L solutions)
Heat change: 5.016kJ (used equation deltaH=m * delta T* Q (which i was given to be 4.18 x 10^-3 kJ/g C

Now i need the amount of NaOH(moles)
and Average H (kj/mol NaOH)

If you're asking for the enthalpy change/mole of sodium hydroxide, simply divide the heat quantity by the number of moles of sodium hydroxide, that will be your answer, in most cases however they want the change in enthalpy per mole of reaction. I'm assuming that you know how to calculate the moles of sodium hydroxide, if you don't...well that would be sad.

GCT said:
If you're asking for the enthalpy change/mole of sodium hydroxide, simply divide the heat quantity by the number of moles of sodium hydroxide, that will be your answer.
Hmm ok...
Well the moles of NaOH... I am confused... the equation is just NaOH so its just 1 mole.
However.. if I use how many moles of NaOH in the 5.5g added itd be 5.5g/39g which is 0.14. Which one is right?

So then itd be... -4.59/0.14=-32.78 kJ/mol
OR -4.59/1=-4.59 kJ/mol
Or have i lost my mind and am completely wrong. I am sorry I should remember this from grade 11 chemistry but I can't, I know that's sad.. but this is why I asked. Last edited:
The molar mass of sodium hydroxide is approximately forty, otherwise you were right in calculating the moles of the compound. You do see that by varying the amount of sodium hydroxide the heat of reaction will differ, that's why it is important to factor in the moles of compound or reaction.

It seems that from your original post that the heat of reaction is 5.016 kJ, be sure to consider the sign. If your teacher actually wants the answer in terms of moles of sodium hydroxide then you can simply divide this value by the moles of sodium hydroxide (you seem to have the right idea here
So then itd be... -4.59/0.14=-32.78 kJ/mol
OR -4.59/1=-4.59 kJ/mol

I don't know where the 4.59 came from. Remember to adjust the moles of sodium hydroxide value.

Thanks

I see now.. thanks soo much (the change in #'s i realized I had the wrong answer that's all...)

good, glad you figured it out.