[natalie_2006]

How many grams of sulphuric acid will neutralize 10.0g of sodium hydroxide?
Then what vloume of water vapour at 100 degrees Celcius and 110 kPA would also be produced?

[mike]

Attempt these yourself first, try starting with the equation for the neutralisation.

[natalie_2006]

Sorry.. I didn't include my answer with it....

So first the molar concentration of sulphuric acid:
is 98.1

NaOH is 40.0

The neutralization equation would be:

1H2SO4  +  2NaOH --> (2Na)+   +  (SO4)-  + H2O


10.0g divided by 40.0 g/mol
= 0.25 g/mol
=2.5 x (10)-1 mol


for the second part the equation is PV=nRT

100*C=(273+100)K
=373 K

110kpa/101.3kpa
=1.08 atm

R = 0.0821

PV = nRT
1.08 * V = 1 * 0.082 * 373
thus V = 0.082 * 373 / 1.08
=28.32 L


Am I right???

[mike]

H₂SO₄ + 2NaOH ---> 2NaSO₄ + 2H₂O

if the number of moles of NaOH = 0.25 moles then the number of moles of H₂SO₄ required to neutralise it must be half of this, ie 0.125 moles.

I'm not sure about the second question as PV=nRT is the ideal gas equation and water vapour is not an ideal gas (not even a gas in fact). Where did you get the value for n in PV=nRT?

[natalie_2006]

Alright I understand the first part now. Thanks.

I got the value for n was when i did the calculation from the first part of the question, the number of moles.  but i believe it's wrong now. Should the value for n be 0.125 moles and then keep the rest of the equation the same?

[mike]

oh I see, so they are part of the same question?

I guess then that you need to know the number of moles of water produced in the reaction. Every one mole of H2SO4 will produce 2 moles of H2O, right? So 0.125moles of H2SO4 will produce 0.25moles of H2O.

P = 110kPa
n = 0.25moles
R = 8.3145 L.kPa.K-1.mol-1
T = 100C = 373K

PV = nRT

therefore:

V = nRT/P = (0.25moles x 8.3145L.kPa.K-1.mol-1 x 373K)/110kPa = 7.05L

I am still not convinved that this is how to work out the answer, someone else might have a better idea.

[mike]

Maybe try looking at saturation vapour pressure

[Borek]

volume of water vapour at 100 degrees Celcius and 110 kPA would also be produced?

Pressure is above standard - so water boils above 100 deg C. There is no reason for it to be in gaseous state.  You may assume there is no water vapour, it will be as good answer as any other.

To be correct one should take into account fact, that every liquid is always in equilibrium with its own vapour, however, such vapour - as every gas does - fills all available space (volume). This available volume is not given in the question, so even knowing that the pressure of saturated vapour over a water at 100 deg C is 1 atm (guess why) you can't answer the question.

[mike]

Pressure is above standard - so water boils above 100 deg C. There is no reason for it to be in gaseous state.  You may assume there is no water vapour, it will be as good answer as any other.

No water vapour? That doesn't make sense does it? I think there would be water vapour where ever there is water, am I wrong. Water vapour and steam are different, steam is a gas (still not ideal).

[Borek]

No water vapour? That doesn't make sense does it? I think there would be water vapour where ever there is water, am I wrong. Water vapour and steam are different, steam is a gas (still not ideal).

Perhaps my English failed me. I am referring to the gaseous water in equilibrium with liquid water. As I explained in the second paragraph of my previous post - there will be gaseous water, but without knowing volume of the tank it is kept in you can't calculate its amount from the data given.

On the second thought - zero volume is the only reasonable answer. If you put water into syringe like tube, and you apply pressure given in the question to the piston, all gaseous water will condense and you will be left with liquid water only.

3:40 am. Good night.