Stoichiometry Help-chem limiting reactants

For this equation 2NaCl+H2SO4=2HCL+Na2SO4

What is the excess of H2SO4 left? If 5.06 of each reactant is taken.

I know NaCl is the limiting reagent, so there will be 0 of that left since all of it reacts, but I'm not sure how to get the left over of H2SO4.

I found the moles of H2SO4 present, which is .051591 (5.06/98.079)
I also found the moles of NaCl present, which is .086585 (5.06/58.44)

The problem I'm encountering is that .051591-.086585 is a negative number and my answer cannot be negative.

The coefficients in the balanced equation determine the ratio of moles of reactants and products.
For this equation 2NaCl+H2SO4=2HCL+Na2SO4
2 moles of NaCl react with 1 mole of H2SO4 to produce 2 moles of HCl and 1 mole of NaSO4
Number of moles of NaCl = 5.06 ÷ 58.5 ≈ 0.0865
Number of moles of H2SO4 = 5.06 ÷ 98.1 = 0.05158

Since the mole ratio of NaCl to H2SO4 is 2 : 1, the number of moles of H2SO4 = ½ * number of moles of NaCl.

Number of moles of H2SO4 = ½ * 5.06 ÷ 58.5 ≈ 0.04325
This is the number of moles of H2SO4 which reacted with NaCl.

Excess H2SO4 = (5.06 ÷ 98.1) – (½ * 5.06 ÷ 58.5) = 0.00833 moles

Mass of excess H2SO4 = 98.1 * [(5.06 ÷ 98.1) – (½ * 5.06 ÷ 58.5)] = 0.817 grams.
The mass of excess H2SO4 is approximately 0.817 grams!