[halp]

Hi, my question is in the link below:
https://www.dropbox.com/s/vr99iv1u9k56q3p/dd.png?dl=0

My attempted answer is below:

I'm struggling to find the mass. Is the mass 50.0mL + 25.0mL = 75.0mL solution
It says to assume 1mL solution = 1g solution
Is this referring to the solution or just the water?
If it is referring to the solution, would the mass be 75.0g?

q(soln)=m(soln)cΔT
=75.0 x 4.184 x (27.21-25)
=693.5J
=0.6935kJ

q(rxn)= -0.6935kJ
ΔH=-0.6935kJ

I also struggled to find the moles. What moles do I find? the NaOH or HCl? I wasn't sure, but I just used the HCl.

n(HCl)=0.025L x 0.5M
=0.0125mol

ΔH=-0.6935kJ/0.0125mol
=-55.48kJ/mol

Could you help me understand how to find the mass and which moles to use? Thank you

[Borek]

If it is referring to the solution, would the mass be 75.0g?

Yes.

I also struggled to find the moles. What moles do I find? the NaOH or HCl? I wasn't sure, but I just used the HCl.

This is a limiting reagent type problem.

[halp]

Thanks.
I didn't limiting reagents could be used in enthalpy type questions!

[halp]

One more question, what significant figures should i use for my calculations?
I think that i have to use the least precise figure. so would it be 3 significant figures due to 75.0g?

[halp]

Sorry, i thought i wouldnt have any more questions about this, but I talked to my tutor and he said the moles isnt HCl. He said to use moles of water.
I showed him a website to say that heat of solution is energy absorbed/released when solvent is dissolved in solution.
http://www.ausetute.com.au/heatsolution.html
What he said is below:
"It is per mole of water produced from what I know,remember this is a neutralisation reaction.
That is a different case,it is a case where a solute is dissolving exothermically in water,which is not our case here.
Our case is where two chemicals are reacting resulting in neutralisation,so this is enthalpy change of neutralisation,the one you showed me is enthalpy change of solution.
A base + acid=enthalpy change of neutralisation.
I see the part that that you were talking about,in this case the enthalpy of solution=neutralisation"

Could someone clarify please? I'm so confused now. So is it moles of HCl or H20?

[DrCMS]

A good place to start is with a reaction equation.

[halp]

NaOH (aq) + HCl(aq) --> NaCl(aq)+H2O(l)

n(NaOH)=0.025mol
n(HCl)=0.0125mol
HCL is the limiting reagent

but my tutor said to use the mole of water because we are dealing with a neutralisation reaction?
should i use mole of HCl or water...

[DrCMS]

OK how many moles of water are formed?

[halp]

n(H20)=0.0125mol

ΔH=-0.6935kJ/0.0125mol
=-55.48kJ/mol

This answer is the same as using moles of HCl, but would it be more right to use moles of HCl or water? the question doesnt specify if they want the enthalpy of HCL, or water.

[DrCMS]

What would have happened if you had twice as much HCl or three times as much?

[halp]

the ΔH would double or triple?

[KungKemi]

In order to determine the molar heat of neutralisation one needs to look at the moles of produced water, since this is directly related to the activity of hydrogen and hydroxide ions in solution. In this case, the molar heat of solution for this reaction is the same as the molar heat of neutralisation, however, for other cases one could simply find the enthalpy of solution by looking at the dissociated reactant/s.

KungKemi

[DrCMS]

the ΔH would double or triple?

Yes and no.

With the amount of sodium hydroxide listed double the HCl would give double the total heat output but identical heat per mole of HCL used or water produced.
With the amount of sodium hydroxide listed triple the HCl would only give double the total heat output and the same heat per mole for water produced as above but the wrong answer if you used the total moles of HCL added.

[halp]

Alright, so should i use the moles of the limiting reagent which is HCl or the moles of water produced in this question?

[XeLa.]

In this case, you can use the moles of HCl since it is the limiting reagent, and it is in a 1:1 ratio with the produced water. However, just in future if you are asked to find the molar heat of neutralisation for a reaction then always use the produced moles of water as the reference.