[funboy]

Q) How many grams of sulphuric acid will neutralize 10.0g of sodium hydroxide


A)

so the balanced equation

H2SO4 + 2NaHO ----> Na2SO4 + 2H20

Molar mass of H2SO4 = 98.1 g/mol
Molar mass of NaHO = 40.0 g/mol

10.0/40.0 = .25 mol

H2SO4 and 2NaHO have a 1:2 ratio

 x        .25
----  = -----
 1         2


x = .125

.125 X 98.1 = 12.263

Therefore 12.3 g of sulphuric acid is required to neutralize 10.0 g of sodium hydroxide

[xiankai]

everything is correct, but for a nitty gritty detail that i'll run you through

in all steps of your calculation, remember to round to 3 significant figures, so that .25 mol becomes .250 mol.

apart from that, well done.

PS: you can use the 'sup' or 'sub' icons while posting to make the numbers something like this: ² ₂

[Neecze]

I thought chemical formula of 'sodium hydroxide' is NaOH.

[Alberto_Kravina]

I thought chemical formula of 'sodium hydroxide' is NaOH.

NaOH = NaHO, although NOBODY (maybe except funboy ) writes Sodium hydroxide as "NaHO".

[funboy]

Its 7am and Im just finishing up my 2nd brake at work (yea grave shifts are fun fun fun).

The second part to this question is

what volume of water vapour at 100C and 110kPa would also be produced.  I havent seen
a question like this before so its expected that I actually learned something in the lessons and can apply it.  This of course leaves me uncertain.

because the question involves Volume and the temp and pressure are given to me I will need to use
PV=nRT

so V = (nRT/P)

n = .125
P = 110kPa
T = 373 K
R = 8.31 kPa L

so V = (.125 X 8.31 X 373) / 110

V = 3.522 L or water vapour

Am I anywhere close to doing what I should be to answer this question??

[Borek]

V = 3.522 L or water vapour

Am I anywhere close to doing what I should be to answer this question??

You are off by the factor of 2. Look at the reaction equation and compare number of moles of H₂O with the number of moles of NaOH. Otherwise you are on the right track.

[funboy]

when comparing the number of moles between NaOH and H2O they both have 2 moles.

so n = .250

Im slightly confused about why Im focusing on NaOH

I understand that the hydroxide ion and the hydrogen ion combine to create water, buy why focus on NaOH and not H2SO4.

[xiankai]

u can focus on H₂SO₄ or NaOH, it makes no difference since u're ultimately comparing molar ratios. just don't forget when focusing on H₂SO₄, you multiply it by 2 based on the mole ratio of course 

[Borek]

when comparing the number of moles between NaOH and H2O they both have 2 moles.

While this is not an incorrect way of stating things, I would rather state they both have the same number of moles.

I understand that the hydroxide ion and the hydrogen ion combine to create water, buy why focus on NaOH and not H2SO4.

Number of moles of NaOH was (indirectly) given in the question, thus it is easier to base your calculations on NaOH, not H₂SO₄.

[skyglow1]

everything is correct, but for a nitty gritty detail that i'll run you through

in all steps of your calculation, remember to round to 3 significant figures, so that .25 mol becomes .250 mol.

apart from that, well done.

PS: you can use the 'sup' or 'sub' icons while posting to make the numbers something like this: ² ₂

Are you sure about this? We were taught to only round to 3sf at the very end. If you round at each step, the roudning error would be too large. Its only your final answer that cannot be more accurate than 3sf. Or am I wrong?

[Borek]

in all steps of your calculation, remember to round to 3 significant figures, so that .25 mol becomes .250 mol.

Are you sure about this? We were taught to only round to 3sf at the very end. If you round at each step, the roudning error would be too large. Its only your final answer that cannot be more accurate than 3sf. Or am I wrong?

To clarify: you should never round intemediate results used for further calculations, but you should always round them if you are going to show them.

For example: molar mass of NaOH is 39.9971 (6sf). We have 10.0 g of NaOH (3sf). My calculator displays 0.2500181 mole as a result of division - that's the number I will use for further calculations, but I am not going to tell there is 0.2500181 mole of NaOH - there is only 0.250 mole

[skyglow1]

Ah yes, that is what we do in maths too, storing the more accurate value and using it while writing down the rounded value. Thanks