Question

If 1 grams of sodium bicarbonate are decomposed, determine the combined mass loss (in grams) of water and carbon dioxide that are produced. When a 1.08g mixture of sodium chloride and sodium bicarbonate were heated for several minutes in a crucible and the mass after the first 3 heating intervals remained 0.8 grams.

What is the combined mass loss of water and carbon dioxide per gram on sodium bicarbonate?

Answer 1

Ans. 1. 2 NaHCO₃(s) -----Heat---------> Na₂CO₃(s) + CO₂(g) + H₂O(g)

Moles of NaHCO₃ = Mass/ Molar mass’

                                    = 1.000 g / (84.006908 g/ mol)

                                    = 0.011904 mol

According to the stoichiometry of balanced reaction, 2 mol NaHCO₃ produces 1 mol each of CO₂ and H₂O.

So,

            Moles of H₂O = moles of CO₂ = (1/2) x Moles of NaHCO₃

                                                            = (1/2) x 0.011904 mol

                                                            = 0.005952 mol

Now,

Total loss in mass = Mass of 0.005952 mol CO₂ + Mass of 0.005952 mol H₂O

                        = (0.005952 mol x 44.0098 g mol^-1) + (0.005952 mol x 18.01528g mol^-1)

                        = 0.261942 g + 0.107225 g

                        = 0.369167 g

#2. # Loss in mass after 4th drying = Initial mass of sample – Dried mass of sample

= 1.08 g – 0.8 g

= 0.28 g

# Calculate Moles of NaHCO₃ in sample:

H₂O and CO₂ are lost as gases. So, net loss in mass during drying must be equal to the mass lost in form of H₂O and CO₂.

See the stoichiometry of decomposition, 2 mol NaHCO₃ produces releases 1 mol each of CO₂ and H₂O.

Let the number of moles of NaHCO₃ in the mixture be X.

So,

Moles of H₂O lost = 0.5X

Moles of CO₂ lost = 0.5X

Now,

Mass of H₂O lost = Moles x Molar mass = 0.5X mol x (18.01 g/mol) = 9.005X g

Mass of CO₂ lost = 0.5X x (44.01 g/mol) = 22.005X g

Total mass lost in form of H₂O and CO₂ = 9.005X + 22.005X = 31.010X g

So, 31.010X g must be equal to 0.28 g – the mass lost during drying.

Or, 31.010X g = 0.28 g

Or, X = 0.28 / 31.010 = 0.009029

Hence, X = 0.009029

Therefore, moles of NaHCO₃ in sample = X moles = 0.009029 mol

# Calculate mass of NaHCO₃ in original sample:

Mass of NaHCO₃ in sample = Moles of NaHCO₃ x Molar mass

= 0.009029 mol x (84.006908 g/mol)

= 0.7585 g

# Combined mass lost = 0.28 g

Mass of NaHCO₃ in sample = 0.7585 g

Now,

Combined mass loss per g Na-bicarbonate = Combined mass loss / Mass of NaHCO₃

                                                = 0.28 g mass loss / 0.7585 g NaHCO₃

                                                = 0.369 g CO₂ and H₂O per gram NaHCO₃