2 L2 L of 0.2MH2SO40.2MH2SO4 is reacted with 2 L2 L of 0.1MNaOH0.1MNaOH solution, the molarity of the resulting product Na2SO4Na2SO4 in the solution is____millimolar.

To find the molarity of the resulting product Na2SO4 in the solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between H2SO4 and NaOH.

The balanced chemical equation for the reaction is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH to produce 1 mole of Na2SO4.

Given that the initial volume of H2SO4 solution is 2 L and its molarity is 0.2 M, we can calculate the number of moles of H2SO4 using the formula: moles of H2SO4 = volume of H2SO4 solution (in L) x molarity of H2SO4

moles of H2SO4 = 2 L x 0.2 M = 0.4 moles

Since the stoichiometry of the reaction is 1:1 between H2SO4 and Na2SO4, the number of moles of Na2SO4 produced will also be 0.4 moles.

Now, we need to find the molarity of Na2SO4 in the solution. To do this, we divide the number of moles of Na2SO4 by the final volume of the solution.

The final volume of the solution is the sum of the initial volumes of H2SO4 and NaOH solutions, which is 2 L + 2 L = 4 L.

Molarity of Na2SO4 = moles of Na2SO4 / volume of solution (in L) Molarity of Na2SO4 = 0.4 moles / 4 L = 0.1 M

To convert the molarity to millimolarity, we multiply by 1000: Molarity of Na2SO4 = 0.1 M x 1000 = 100 millimolar

Therefore, the molarity of the resulting product Na2SO4 in the solution is 100 millimolar.