How to find the molarity of balanced equation ?

H₂SO₄(aq) + 2 NaOH(aq) → 2 H₂O(ℓ) + Na₂SO₄(aq)
Mole ratio H₂SO₄ : NaOH = 1 : 2

Moles of NaOH = (0.2401 mol/L) × (16.44/1000 L) = 0.0039472 mol
Moles of H₂SO₄ = (0.0039472) × (1/2) = 0.0019736 mol
Molarity of H₂SO₄ = (0.0019736 mol) / (20.00/1000 L) = 0.09868 M

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OR:
(0.2401 mol NaOH / 1000 mL NaOH solution) × (16.44 mL NaOH solution) × (1 mol H₂SO₄ / 2 mol NaOH) / (20.00/1000 L H₂SO₄ solution)
= 0.09868 M H₂SO₄

Find the number of moles of NaOH:

16.44 mL x 0,2401 mol/L  = 3.947 m mol

Then divide by the mL of H2SO4 after diving in half to account for getting 2 H+ per mole:

3.947 m mol / 20.00 mL /2 = 0.09868 mol/L

.You can ignore the added water, it does not change the moles of reactants/

some details to note... 
.. (1) you want molarity H2SO4 = mol H2SO4 / L H2SO4 solution
.. (2) mol / L = mmol / mL
.. (3) we're going to set this up and solve it via dimensional analysis
.. .. . .with proper sig figs!

here's the solution

16.44mL NaOH.. ..0.2401 mmol NaOH... 1 mmol H2SO4
----- ----- ---- ----- x ----- ----- ----- ---- ----- x ----- ----- ---- ----- = 0.09868M H2SO4
20.00mL H2SO4... ..... 1 mL NaOH... ... ... 2 mmol NaOH

************
notice how 
.. mL NaOH / mL NaOH
.. mmol NaOH / mmol NaOH
cancel out leaving
.. mmol H2SO4 / mL H2SO4 = molarity H2SO4?

the calcs are simple
.. 16.44 / 20 * .2401 / 2 = 
and then you round to 4 sig figs.

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Bobby_notsothinanymore

I read your comment and get it that YOU don't understand my solution.  Let me help you out.

Since about......oh.... 1980 or so, chemistry teachers have been teaching "factor label method" (which is often called "dimensional analysis") to chemistry students. Why?  For loads of reasons including
.. (1) it's a consistent and reliable approach to solving problems
.. (2) it has 2 built in error checks
.. (3) it's easy to track sig figs
.. (4) it identifies missing information that you need to look up or RESEARCH
.. (5) it's more precise (no intermediate rounding = better precision)
.. (6) it's FAST and nearly effortless.
.. (7) it ALWAYS leads to better grades.  Never the other way around

So what are the basics?  5 principles
.. (1) any # with units * 1 = that same intrinsic value.. 12in * 1 = 12in
.. (2) units cancel.  Just like 2 / 2 cancels out, cm / cm cancels out
.. (3) for any equality a = b, we can write a / b = b / a = 1
.. (4) units can be converted by combining (1), (2), and (3)
.. (5) the power rule.  since 1^n = 1, if a=b, any (a/b)^n = (b/a)^n = 1

let's me give you an example.  let's say I want to convert 60 miles to feet
.. 60 mi * 1 = 60mi.. . .... ... .. ... .... ... ... ... ... ... ... .. ... .. principle (1)
.. 5280ft = 1 mi ---> (5280ft / 1 mi) = (1 mi / 5280ft) = 1.. principle (3)
.. 60mi * (5280ft / 1 mi) = 60mi...  ... ... ... ... .. ... ... ... ... .. .subbing
.. 60 * 5280ft = 60mi...  . ... ... ... ... ... ... ... ... ... ... .. . .. ..principle (2)
.. 316800ft = 60mi
now in practice we don't do all the subbing, we just start with what we want to convert FROM on the left and convert to the right
.. 60 mi * (5280ft / mi) = 316800ft
notice I'm simply multiplying 60mi * 1?  that doesn't change the intrinsic value, but does change the units

So what is the best way to apply this technique?
.. (1) sort through the problem..ID what you're converting FROM
.. .. . what you're converting TO and all the unit factors (those ratios 
... . . .like (5280ft / mi) are called unit factors)
.. (2) start with the FROM on the left and convert to the right
.. (3) convert numerator first, then denominator
.. (4) setup your conversion first, then error check
... . .. .left to right, make sure units cancel
.. . .... right to left, verify conversion factors are correct.
.. (5) run the calcs 2x if time allows
.. (6) identify the factor with the least sig figs.  That is the sig figs in the result
.. (7) any unit factors that you look up must have at least as many
.. . . .sig figs as the value you're converting from.  preferably they should
.. .. . have 1 more

example.  convert 60. mi / hr to m / s using correct sig figs

 60. mi.. 5280 ft... 12 in .. 2.54 cm.. .. 1 m.. . . ... 1 hr
---- ---- x ---- ---- x --- --- x ----- ----- x ---- ---- x ----- ---- --- = 27 m/s
.. 1 hr.. . ..1 mi... .. 1 ft... . .. 1 in... .. .100 cm... 3600 sec

I started with 60. mi/hr on the left (with 2 sig figs)
each fraction is a unit factor = 1
mi / mi.... ft / ft... in / in.... cm / cm.... hr / hr... all cancel out leaving m/s
all the unit factors are correct.. 1hr = 3600s.. 1m = 100cm.. 2.54cm = 1in, etc
the calcs are
.. 60 * 5280 * 12 * 2.54 / 100 / 3600 = 

example of principle (5)
.. 5.5ft³ * (12in / 1ft)³ * (2.54cm / in)³ = 1.6x10^5 cm³
which is calc'd by
.. 5.5 * 12³ * 2.54³ = ?

********
now let's look at the CHEMISTRY problem we were given
.. (1) balanced equation is
.. . . .. 1 H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq)
.. (2) we WANT mmol H2SO4 / mL H2SO4
.. (3) we're converting FROM 16.44mL NaOH and 20.00mL H2SO4
.. (4) our unit factors are
.. .... .. 1 mmol H2SO4 = 2 mmol NaOH (from balanced equation)
.. ... .. . 0.2401 mmol NaOH = 1 mL NaOH (from molarity NaOH)

the rest is simple.  start with the FROM
..16.44mL NaOH
 ---- ---- ---- ---- ---- x 
.20.00 mL H2SO4

convert the numerator..
..16.44mL NaOH .. .. .0.2401 mmol NaOH
 ---- ---- ---- ---- ---- x ---- ---- ---- ---- ----- ---... gives mmol NaOH
.20.00 mL H2SO4.... .. .. 1 mL NaOH

convert the numerator.. again
..16.44mL NaOH .. .. .0.2401 mmol NaOH .. . 1 mmol H2SO4
 ---- ---- ---- ---- ---- x ---- ---- ---- ---- ----- --- x ----- ---- ---- ----- = __M H2SO4
.20.00 mL H2SO4.... .. .. 1 mL NaOH.. ... .... .2 mmol NaOH

and we can see we're left with molarity H2SO4.
the calcs are
.. 16.44 / 20 * 0.2401 / 2 = 
with rounding to 2 sig figs

Bobby_notsothinanymore.....this is what we currently teach chemistry students.  study this.  you can learn it too. 

Here is a complete solution with explanations.

Not a one line calculation which nobody understands  

moles of NaOH at the equivalence point = .2401 moles / liter * 0.01644 L 
= 0.003947244 moles
these react 2:1 with the H2SO4
so moles of H2SO4 at the equivalence point = 0.003947244 /2 = 0.001973622
these are in 20 mL of solution 
so molarity of the H2SO4 = 0.001973622 * 1000/20 = 0.09868 M ( 4 significant figures)  

answer 0.09868 M >>>>> answer