"How many grams of water can be produced when 20.0 mL of 0.600 M H2SO4 react with 40.0 mL of 0.800 M NaOH"

0.432 g

Explanation:

Equation of the reaction;

H2SO4(aq) + 2NaOH > Na2SO4(aq) + 2H2O(l)

Amount of H2SO4= concentration × volume = 20.0/1000 × 0.600 M= 0.012moles

If 1 mole of H2SO4 yields 2 moles of

water

0.012 moles of H2SO4 yields 0.012 × 2/1 = 0.024 moles

For NaOH

Amount= 40/1000 × 0.8 = 0.032 moles

If 2 moles of NaOH yields 2 moles water

0.032 moles of NaOH also yields 0.032 moles of water

Hence H2SO4 is the limiting reactant

Thus;

Mass of water = 0.024 moles × 18g/mol = 0.432 g