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 soluti

The reaction between H2SO4 and NaOH is as follows:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the balanced chemical equation, we can see that one mole of H2SO4 reacts with two moles of NaOH to produce one mole of Na2SO4.

Given that the volume of H2SO4 is 2L and its molarity is 0.2M, the number of moles of H2SO4 is:

Moles = Molarity x Volume = 0.2M x 2L = 0.4 moles

Similarly, the number of moles of NaOH is:

Moles = Molarity x Volume = 0.1M x 2L = 0.2 moles

Since one mole of H2SO4 reacts with two moles of NaOH, the limiting reagent in this reaction is NaOH. Therefore, the number of moles of Na2SO4 produced is half the number of moles of the limiting reagent (NaOH), which is:

Moles of Na2SO4 = 0.2 moles / 2 = 0.1 moles

The total volume of the solution is the sum of the volumes of H2SO4 and NaOH, which is:

Total volume = 2L + 2L = 4L

Therefore, the molarity of Na2SO4 in the solution is:

Molarity = Moles / Volume = 0.1 moles / 4L = 0.025M

So, the molarity of the resulting product Na2SO4 in the solution is 0.025M.