Suppose you spill 45.0 mL of 0.150 M nitric acid on the lab table. You decide to neutralize it with .350 grams of calcium hydroxide

Question

A) Find the limiting reactant: Was the acid completely neutralized?

B)Determine the mass of calcium nitrate that would be recovered from the reaction?

Answer 1

Balanced equation:

2HNO₃ + Ca(OH)₂ ==> Ca(NO3)₂ + 2H₂O

moles HNO3 present = 45.0 ml x 0.150 mmoles/ml = 6.75 mmoles = 0.00675 moles

moles Ca(OH)2 present = 0.350 g x 1 mole/74.09 g = 0.00472 moles

Limiting reactant is nitric acid (HNO₃₎ because you need 2 moles HNO₃ for each 1 mole Ca(OH)₂.

So, yes...the acid is completely neutralized.

How much Ca(NO₃)₂ is recovered?

0.00675 moles HNO₃ x 1 mole Ca(NO₃)₂/2 moles HNO₃ = 0.003375 moles Ca(NO₃)₂ recovered

0.003375 moles Ca(NO₃)₂ x 164.1 g/mole = 0.554 g Ca(NO₃)₂ recovered (to 3 sig. figs.)