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 1.09g 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 0.83 grams.

Answer 1

The balanced equation of decomposition of sodium bicarbonate into sodium carbonate, water and carbon dioxide can be written as follows:

Thus, 2 moles of NaHCO3 gives 1 mole of Na2CO3, 1 mole of H2O and 1 mole of CO2.

Molar mass of NaHCO3 is 84 g/mol, molar mass of Na2CO3 is 106 g/mol, molar mass of H2O is 18 g/mol and molar mass of CO2 is 44 g/mol.

Number of moles of NaHCO3 in 1g can be calculated as follows:

Number of moles of NaHCO3 = 1.0g / (84 g/mol) = 0.012 mol

Thus, number of moles of H2O lost will be 0.012/2 = 0.006 mole and number of moles of CO2 lost will be 0.012/2= 0.006 mole.

Weight of 0.006 moles of H2O = (0.006 mol) x (18 g/mol) = 0.108g

Weight of 0.006 moles of CO2 = (0.006 mol) x (44 g/mol) = 0.264g

Therefore, combined mass loss due to water and CO2 = 0.108g + 0.264g = 0.372g

Hence, from decomposition of 1g of NaHCO3 combined weight loss due to H2O and CO2 is 0.372g

Weight loss from mixture of NaCl and NaHCO3 = 1.09g – 0.83g = 0.260g

Thus, weight of NaHCO3 corresponding to loss of 0.260g = (1g x 0.260g) / 0.372g = 0.699g

Therefore, weight of NaCl in mixture = 1.09g – 0.699g = 0.391g

Hence, 1.09g of mixture contains 0.699g of NaHCO3 and 0.391g of NaCl.