Question

YOU DO:

1) A 15.0 gram sample of sodium hydrogen carbonate, NaHCO3, was contaminated with an impurity. In order
to determine the purity of the sample, it was heated to decompose the material according to the following
reaction:
2NaHCO3 → Na2CO3 + H2O + CO2
If 6.35 grams of sodium carbonate, Na2CO3, were recovered, what percentage (by mass) of the sample was
sodium hydrogen carbonate, NaHCO3?
I

Answer 1

67.1%

Step-by-step explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

Moles Na₂CO₃ - 105.99g/mol-:

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

67.1%