Sulphuric Acid Neutralizing Sodium Hydroxide

I'm trying to complete a homework question. I'll go as far as I've gotten and see if someone can help me farther.

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

A: H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O
Ratio of 1:2

Molar mass of NaOH (23+16+1)=40g 10g/40g=0.25mol

Molar mass of H2SO4 (2+30.1+64)=96.1g

x/1*0.25/2 = 0.125mol

=0.125mol x 96.1g = 12.3g

Now is where I get stuck. The second part of the question asks:
What volume of water vapour at 100C and 110 kPa would also be produced?

So I'll write what I know so far:
P=110kPa
V= ?
n= (I can't figure out)
R=8.31
T=373K

I'm not sure where to pull the n from. My best guess is that it is 2, but I'm not sure. Can someone point me in the right direction so I can complete my homework question.

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