what mass of NaOH would be needed to neutralize exactly 200cm3 of a solution containing 4.9g per dm3 of H2SO4

3 answers

2NaOH + H2SO4 ==> Na2SO4 + H2O
mols H2SO4 = grams/molar mass = 4.9/98 = 0.05
The equation tells you you will need twice that number of mols of NaOH or 2 x 0.05 = 0.1 mol
Then grams NaOH = 0.1 mol x 40 g/mol = 4.0 grams NaOH.
Note that you don't need the 200 cc UNLESS you change H2SO4 to concentration and I didn't do that. Another way of saying that the grams NaOH needed is grams H2SO4 initiall and it makes no difference how much solvent is used.
OOPS. After posting my answer above I re-read the question and is DOES make difference so I'll post a new answer below. Sorry about that.
2NaOH + H2SO4 ==> Na2SO4 + H2O
You have 200 cc H2SO4 containing 4.9 g/L so that is
4.9 g x 200 cc/1000 cc = 0.98 grams H2SO4.
0.98 grams x (1 mol/98 g) = 0.01 mol H2SO4.
So you know you will need 2 x 0.01 = 0.02 mol NaOH and that will be 0.02 x 40 = ? g NaOH.
Check this. Make sure I didn't make an error AND that I read the problem correctly.