What is the percent yield if this reaction produced 55.0g of ethene from 100.0g of ethanol?
Answers
Please Note - The reaction in the question has been omitted, but the equation is still known.
The percent yield of ethene from 100.0 g of ethanol is 90.395 %.
• The chemical equation for the production of ethene from ethanol can be represented as follows :
C₂H₅OH (ethanol) + conc. H₂SO₄ → H₂C = CH₂ (ethene) + H₂O
• From the equation, we see that 1 mole of ethanol produces 1 mole of ethene.
• Now, we have to find the number of moles contained in 100.0 g of ethanol, as given.
Number of moles = Given weight of ethanol / Molar weight of ethanol
• Molar weight of ethanol (C₂H₅OH) = (2 × 12 g) + (6 × 1 g) + 16 g
= 24 g + 6 g + 16 g
= 46 g
• So, number of moles present in 100 g of ethanol = 100 g / 46 g = 2.173
• Since 1 mole of ethanol produces 1 mole of ethene, 2.173 moles of ethanol will produce 2.173 moles of ethene.
• Mass of 2.173 moles of ethene = Moles of ethene × Molar mass of ethene
• Molar mass of ethene (C₂H₄) = (2 × 12g) + (4 × 1g)
= 24 g + 4 g = 28 g
• Therefore, mass of 2.173 moles of ethene = 2.173 × 28 g = 60.844 g
• So, from the above calculations, we find that the theoretical yield of ethene from ethanol is 60.844 g.
• Now, the actual yield of ethene is given as 55.0 g in the question.
• Therefore, the percent yield of ethene can be calculated as :
% yield of ethene = ( Actual yield of ethene / Theoretical yield of ethene) × 100 %
= (55.0 g / 60.844 g) × 100 %
= 90.395 %
∴ The required percentage of yield is 90.395 %.