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To determine how much of the excess reagent will remain after the reaction has been completed, you need to calculate the amount of the excess reagent used in the reaction and subtract it from the initial amount.

In this case, NaOH is the excess reagent since it is not fully consumed in the reaction. We know that the initial amount of NaOH is 0.15 mol. However, we need to figure out how much NaOH is used based on the stoichiometry of the reaction.

From the balanced equation, we can see that 2 moles of NaOH react to form 1 mole of Na2SO4. Therefore, we can set up the following ratio:

2 moles NaOH : 1 mole Na2SO4

Using this ratio, we can calculate the amount of NaOH used in the reaction:

Amount of NaOH used = (0.098 mol Na2SO4) * (2 mol NaOH / 1 mol Na2SO4) = 0.196 mol NaOH

Now, to find the amount of excess reagent remaining after the reaction, we subtract the amount used from the initial amount:

Amount of excess reagent remaining = Initial amount - Amount used
= 0.15 mol - 0.196 mol
= -0.046 mol

Since we can't have a negative amount of a substance, we conclude that there is no excess reagent remaining after the reaction is completed.

Now let's move on to calculating the percent yield.

The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant, which in this case is NaOH. We already know that 0.076 mol of Na2SO4 is produced from NaOH.

To calculate the theoretical mass of Na2SO4, we need to convert moles to grams using the molar mass of Na2SO4:

Mass of Na2SO4 = (0.076 mol Na2SO4) * (142 g/mol Na2SO4)
= 10.792 g

The percent yield is then calculated by dividing the actual yield (10.5 g) by the theoretical yield (10.792 g) and multiplying by 100:

Percent yield = (10.5 g / 10.792 g) * 100
= 97.09%

Therefore, the percent yield of Na2SO4 is approximately 97.09%.