16.8 g of NaHCO3 when heated produces 9 g of Na2CO3 according to the following reaction:

Question:

16.8 g of NaHCO₃ when heated produces 9 g of Na₂CO₃ according to the following reaction:

2NaHCO₃ Na₂CO₃ + H₂O + CO₂

i. Find the % purity of NaHCO3

ii. How much volume of CO₂ will be produced at 1 atm and 27°C?

Solution:

Balanced equation:

2NaHCO₃(s) → Na₂CO₃(aq) + H₂O(l) + CO₂(g)

a). Calculation of purity of NaHCO₃:

Mole ratio of NaHCO₃ : Na₂CO₃ = 2 : 1.

Molar mass of NaHCO₃ = 84.007 g/mol

Molar mass of Na₂CO₃ = 105.9888 g/mol

Mol of NaHCO₃ available = 16.8 g/(84.007/mol) = 0.2 mol

Mol of Na₂CO₃ produced = 9 g/(105.9888 g/mol) = 0.085 mol.

From the above equation, 0.085 mol of Na₂CO₃ is originated from (2/1) x 0.085 mol of NaHCO₃ = 0.17 mol of NaHCO₃.

Mass of NaHCO₃ reacted = 0.17 mol x 84.007 g/mol = 14.28 g.

Hence, purity of NaHCO₃ = (14.28 g/16.8 g) x 100% = 85.007% by mass.

b). Calculation of volume of CO₂ produced at 1 atm and 27° C:

As mentioned above that NaHCO₃ reacted = 0.17 mol, so, mol of CO2 produced = ½ x 0.17 mol = 0.085 mol (n).

Since CO₂ is an ideal gas, then apply the following formula to calculate the volume of CO₂.

PV = nRT

V = nRT/P

V = (0.085 mol x 0.0821 L.atm/mol.°K x 300°K) ÷ 1 atm

Solving for V, we get,

V = 2.09 L.

So, volume of CO₂ produced = 2.09 L.