Question

When NaHCO3 completely decomposes, it can follow this balanced chemical equation:

2NaHCO3 → Na2CO3 + H2CO3


Determine the theoretical yields of each product using stoichiometry if the mass of the NaHCO3 sample is 3.55 grams. (Show work for both)
In an actual decomposition of NaHCO3, the mass of one of the products was measured to be 2.16 grams. Identify which product this could be and justify your reasoning.
Calculate the percent yield of the product identified in part B. (Show your work)

Answer 1

The balanced equation for the decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water is: 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) Like most chemical reactions, the rate of the reaction depends on temperature.

Answer 2

When NaHCO3 completely decomposes, it can follow this balanced chemical equation : 2NaHCO₃ => Na₂CO₃ + H₂CO₃

We have to determine the theoretical yields of each of the products using stoichiometry if the mass of the NaHCO₃ sample is 3.55 g.

  mass of NaHCO₃ = 3.55 g

  molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.55/84 =  0.04226 mol

you see, two moles of sodium bicarbonate produce one mole of sodium carbonate and one mole of hydrogen carbonate.

so, the no of moles of sodium carbonate = 0.04226/2 = 0.02113 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar masss of Na₂CO₃

=  0.02113 × 106 ≈ 2.24 g

no of moles of hydrogen carbonate = 0.04226/2 = 0.02113 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.02113 × 62 g = 1.31 g

mass of one of the products was measured to be 2.16 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.24 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.16 g

and theoretical yield of Na₂CO₃ = 2.24 g

∴ % yield = 2.16/2.24 × 100 ≈ 96.43 %

Therefore the percentage yield of the product is 96.43%