how many moles of excess reactant are present when 3.20 x 10^2 mL of 0.226 M sulfuric acid reacts with 0.531 L of 0.185 M sodium hydroxide to form water and aqueous sodium sulfate
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
First we identify the limiting reagent(LR).
mols H2SO4 = M x L = about 0.0732
mols NaOH = 0.0982
Convert mols H2SO4 to mols Na2SO4. That will be 0.0732.
Convert mols NaOH to mols Na2SO4 which will be (1/2)*0.0932 or about 0.05 but you need to be more accurate on ALL of these calculations.
You can see that we have two different values for mol Na2SO4 which means one of them must be wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. The other reagent is the excess reagent (ER). So NaOH is the LR and H2SO4 is the ER.
Now that we have identified the LR and the ER, we convert mols LR to mols ER and we do it with coefficients just as in the part above.
about 0.05 x (1 mol H2SO4/2 mols NaOH) = about 0.025 mols. Again, this is an estimate so you need to work through from the beginning to get a good number.