How many moles of which reactant are in excess when 359.0 mL of 0.259 M sulfuric acid reacts with 0.510 L of 0.191 M sodium hydroxide to form water and aqueous sodium sulfate?

Answer 1

Balanced equation first.

2NaOH + H2SO4 -> Na2SO4 +2H2O

Molarity = moles of solute/Liters of solution

0.259 M H2SO4 = moles H2SO4/0.359 liters

= 0.09298 moles H2SO4

0.191 M NaOH = moles of NaOH/0.510 liters

= 0.09741 moles NaOH ( Let us find limiting reactant )

0.09298 moles H2SO4 (1 mole H2SO4/2 mole NaOH)

= 0.04649 moles H2SO4-- Sulfuric acid limits and,

0.04648 moles of sodium hydroxide are in excess.

Answer 2

Balanced equation.

2NaOH + H2SO4 --> Na2SO4 + 2H2O

5 moles NaOH (1 mole Na2SO4/2 moles NaOH)

= 2.5 moles sodium sulfate resultant

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