Quick Question

benzenee

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(edited 11 years ago)

Reply 1

charco

Study Forum Helper

Original post by benzenee

Our experiment was to measure enthalpy changes, I managed to do the first section of it which was solvation -
Q = C M Delta T, which for me was 4.184 * 70g (this is of distilled water)
* 6 (temp change) = 1757.28
then moles = 1.72 grams of NaOH / 40g/mol = 0.043mol
Then determined enthalpy - 1.75728 / 0.043 = 40.867

Now we then placed 70ml of 1.0 HCL acid into calorimeter, measured temp of NaOH from first experiment, when same temp and cool down, then added it to the HCL, so now to determine Q, it is still 4.184 * is it 140g as both liquids are added and m is mass of water in calorimeter, or just the 70g from the first part?

Final Question, how do i figure out the moles, as i did 70g of the solution / 40 = 1.75 mol, but it means it has increased, when no more NaOH has been added, so should i stick with the original amount of 1.72g / 40 = 0.043 or change to 70ml including NaOH and H20?

Thanks

Both liquids are being heated up so it must be 140g

The maximum number of moles is limited by the limiting reagent. As they react together in a 1:1 ratio then this is the component with the fewer moles:

If 0.043 moles of NaOH have been used and 70ml of 1.0 M HCl ( = 0.07 mol) then NaOH is the limiting reagent.

Your energy must be divided by 0.043 ....