How many ml of 250 M H2S04 are required to completely react with 25.0 ml of 150 M NaOH?

The molarities given in this problem are not realistic. NaOH has a molar solubility of about 17 moles per liter at 25ºC, and concentrated sulfuric acid is about 18M. I'll explain the process for solving the problem then you can substitute the correct molarities if those given are incorrect.

This is a neutralization problem. When the NaOH is completely neutralized the moles of acid (H+) and moles of base (OH-) reacted will be equal. Since molarity (M) is moles of solute per liter of solution, moles is molarity times liters:

M = moles/liter and moles = (molarity)(liters)

One factor that we must consider is that sulfuric acid, H2SO4, is diprotic. That means it produces two moles of H+ per mole of H2SO4:

`H_2SO_4(aq) + 2NaOH_((aq)) -> Na_2SO_4_(aq) + 2H_2O`

The number of moles of H2SO4 required for neutralization is equal to half the number of moles of NaOH, or:

2(molarity)(volume) of acid = (molarity)(volume) of base

(dividing both sides by 2 shows that moles acid = (1/2) moles base)

The volume doesn't have to be in liters as long as the same volume unit is used for both solutions. 

Now we'll rearrange to solve for ml of H2SO4:

ml of H2SO4 = (NaOH molarity)(ml of NaOH)/(H2SO4 molarity)

 = (150 M)(25.0 ml)/(250 M) = 15 ml H2SO4