For

H2SO4 + 2NaOH = 2H2O + Na2SO4

n(H2SO4) = 9.65g/98g/mol = 0.098
n(NaOH) = 6.10g/40g/mol = 0.15 mol

1 mole of H2SO4 is = to 1 mole of Na2SO4. H2SO4 yields 0.098 mol of Na2SO4.
2 moles of NaOH is = to 1 mole of Na2SO4. NaOH yields 0.076 moles of Na2SO4.

The limiting reactant is NaOH

How much of the excess reagent will remain after the reaction has been completed?

And

If 10.5-g of Na2SO4 are actually recovered experimentally, what is the percent yield?