[asmcriminal]

 “Calculate the heat of solution per mole of NaOH”

The data I have is:
2.12g NaOH(s)
H2O = 50ml
Mass of solution: 52.12g
ΔT = 9.2997°C
Heat Capacity of 1.0M NaOH is 3.9 J/g°C



In the lab manual it gives an example.

“If the solution has a mass of 52.0g and its heat capacity is 4.1 J/g°C, then the heat change is calculated as follow
q = mass * specific heat * temperature
q = 1.73 * 10^3 J or -1.73kj(exothermic)
If 0.052 moles of product is formed, the molar enthalpy change is
-1.73kj/0.052 mol = -33 kj/mole)”

[Dan]

Please post what the experiment is and what you are trying to calculate.

The best thing to do is post the question exactly as it is given to you.

[asmcriminal]

Please post what the experiment is and what you are trying to calculate.

The best thing to do is post the question exactly as it is given to you.

Dan,

Thanks, i edited the original post.

[Dan]

You have an equation for q, so try plugging in your numbers.

[asmcriminal]

I did all that, I am trying to find the "moles of product formed"


q = mCΔT = 102g * 3.9J/g°C * 9.2997°C = 3699.42K or3.7 kJ

now the step is to divide the kj by products formed...
NaOH(s) ---> Na+(aq) + OH-(aq)

I have to divide 3.7kj by the moles of product formed to get the molar enthalpy

So how many moles of product are formed?

[Dan]

I did all that, I am trying to find the "moles of product formed"


q = mCΔT = 102g * 3.9J/g°C * 9.2997°C = 3699.42K or3.7 kJ

Where did 102 g come from? You wrote 52.12 g for the mass in your original post.

So how many moles of product are formed?

If you have 1 mol NaOH(s), mow many mol of NaOH(aq) do you get?
How many mol of NaOH(s) did you start with?

[asmcriminal]

I didn't get the notifications to my email, well i did... but it didn't notify to my phone for some reason. I know this post is old. I am just posting this for closure, incase someone else needs help with a similar problem. This is what I wrote, and my professor told me it was okay.

 Total volume of HCL(aq)+NaOH(aq) = 100.2mL

Density of NaCl Solution = 1.02g/mol

100.2mL(1.02/1 mol ) = 102g (Mass Of Solution)

q = m*C*DeltaT

m = 102g
C = 4.0 J/g(degrees)C
DeltaT = 6.53346(degrees)C

102g * 4.0J/g(degree)C * 6.53346 = 2665.65J or 2.7kj

DeltaH = 2.7/moles(of product(NaCl))

Solutions used 1.1289M HCL 50.1mL  and 50.1mL Of 1.0202M NaOH
NaOH is the limited reacting it has 0.051112 mols

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

0.051112 mols NaOH(1mol NaCl/1mol NaOH) = 0.051112 mols of NaCl