Question

How many moles of water is produced from the reaction of 15.0 ml of 2.00 M sulfuric acid with 13.5 ml of 2.00 M potassium hydroxide?

Answer 1

Let us write Balanced equation:
H₂SO₄ + 2 KOH ===> K₂SO₄ + 2 H₂O
Reaction type: double replacement

Volume of H2SO4 = 15.0 ml

Concentration of H2SO4 = 2.00 M

Moles of H2SO4 = 15 x 2 /1000 = 0.03 Moles

Volume of KOH = 13.5 ml

Concentration of KOH = 2.00 M

Moles of KOH = 13.5 x 2 /1000 = 0.027 Moles

Limiting reagent is KOH

Moles of water produced = 0.027 Moles

Balanced equation: H2SO4 + 2 KOH = K2SO4 + 2 H20 Reaction type: double replacement ) Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight H2SO4 98.07848 0.01350000 1.324059 KOH 56.10564 0.027 1.514852 K2SO4 174.2592 0.01350000 2.352499 H20 18.01528 0.02700000 0.4864126 Units: molar mass - g/mol, weight - g.