How many ml of .80M H2SO4 are required to completely neutralize 230mL of .80M Rubidium Hydroxide? Show work?

So we're using a polyprotic strong acid and a monoprotic strong base. Going into this, you know that 1 mol of H+ will react with 1 mol of OH-, and since these are both strong, we ume complete dissociation. First calculate the number of moles of OH- we need to neutralize. We use M=mol/L, so .80=mol/.230. This gives .184mol OH-. This means we need .184mol H+ to completely neutralize the RbOH. Since there are two moles of H+ for every mole of H2SO4, we divide this by 2 to get the number of moles of H2SO4 we need. Needing .092mol H2SO4, we plug this back into M=mol/L, giving .80=.092/L. This gives us .0736L, or 73.6mL. Using sig figs (2) our answer is 74mL.