Question

A student decides to set up an experiment that night before performing the experiment. A 0.40 g sample of sodium hydroxide in a 250 mL beaker that has a mass of 112.58 g. The beaker is allowed to stand open (exposed to air) for 24 hr. At that time, the student observes that the beaker contains a dry, white residue and the total mass of the beaker and the residue is 113.09 g. He proposes that a reaction must have occurred between the sodium hydroxide and the carbon dioxide in the air. The possible reactions are:

sodium hydroxide(s) + carbon dioxide( g) -> sodium carbonate (s)

sodium hyroxide (s) + carbon dioxide (g) -> sodium carbonate (S) + water (g)

sodium hydroxide (s) -> sodium oxide (s) + water (g)

Write the balanced chemical equations for the above reactions.

Determine the amount of solid product expected in each reaction.

Based on these calculations, which of the above reactions would you predict is correct? Show all the work including chemical equations and explain your choice. Thank you.

Answer 1

(1) sodium hydroxide(s) + carbon dioxide( g) ---> sodium carbonate (s)

This reaction is not possible because sodium hydroxide and carbon dioxide react to produce sodium carbonate and water.

However, sodium hydroxide reacts with carbon dioxide to form sodium hydrogen carbonate also known as sodium bicarbonate.

The balanced equation of this reaction is :   NaOH(s) + CO₂ (g) ---> NaHCO₃ (s)

Mass of NaOH used = 0.40 g

Moles of NaOH = 0.40g / 40 g/mol = 0.01 mol

Moles of NaHCO₃ produced = 0.01 mol ( Because NaOH and NaHCO₃ mole ratio = 1:1)

Mass of NaHCO₃ = ( 0.01 x 84) g = 0.84 g

(2) sodium hyroxide (s) + carbon dioxide (g) -> sodium carbonate (s) + water (g)

The balanced equation is: 2 NaOH (s) + CO₂ (g) = Na₂CO₃(s) + H₂O (g)

Moles of NaOH used = 0.01 mol

Moles of Na₂CO₃ produced = 0.01/2 mol = 0.005 mol [ Since mole ratio of NaOH and Na₂CO₃ = 2 : 1]

Mass of Na₂CO₃ = ( 0.005 x 105.98) g = 0.53 g

(3) sodium hydroxide (s) -> sodium oxide (s) + water (g)

The balanced equation is: 2 NaOH (s) = Na2O (s) + H2O(g)

Moles of NaOH used = 0.01 mol

Moles of Na₂O produced = 0.005 mol [ Since mole ratio of NaOH and Na₂O = 2: 1]

Mass of Na₂O produced = ( 0.005 x 61.97) g = 0.31 g

According to the problem, the mass of the white product produced by the reaction

= 113.09g - (112.58 + 0.40)g = 0.11 g

The mass of the product produced in the third reaction = 0.31 g and it is the nearest to the mass 0.11g,  for this reason, it is the correct reaction.