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