Theoretical yield = 2.397
The product could be sodium carbonate
percent yield = 98.456%
When nahco3 completely decomposes, it can follow this balanced chemical equation:
2nahco3 → na2co3 h2co3
If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.
mass of NaHCO₃ = 3.80 g
molar mass of NaHCO₃ = 84 g/mol
so the no of moles of NaHCO₃ = 3.80/84 = 0.0452 mol
You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.
so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol
∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃
= 0.0226 × 106 ≈ 2.397 g
no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol
mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃
= 0.0226 × 62 g = 1.401 g
mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.
percentage yield = experimental yield/theoretical yield × 100
here experimental yield of Na₂CO₃ = 2.36 g
and theoretical yield of Na₂CO₃ = 2.397 g
∴ % yield = 2.36/2.397 × 100 ≈ 98.456%
Therefore the percentage yield of the product is 98.456%
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