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How many moles of excess reactant are present when 3.50 x 10^2 mL of 0.216 M sulfuric acid reacts with 0.510 L of 0.202 M sodium hydroxide to form water and aqueous sodium sulfate?

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Moles H2SO4 = M x L
moles NaOH = M x L
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
moles NaOH = 2 x moles H2SO4
Excess = larger - smaller.
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