How many grams of water can be produce from the complete reaction of excess nitric acid and 33.2 mL of 0.245 M lithium hydroxide solution, assuming that lithium hydroxide is the limiting reactant?

Mass of water produced : 0.146 g

Given

33.2 mL of 0.245 M lithium hydroxide

Required

mass of water

Solution

Reaction

HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)

mol LiOH :

= M x V

= 0.245 x 33.2 ml

= 8.134 mmol

From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol

mass H₂O :

= mol x MW

= 8.134 x 10⁻³ mol x 18 g/mol

= 0.146 g