Question

in the reaction 2C₈H₁₈ + 2SO₂ give as 16Co₂ +18H₂0 when 52.7g of octane burns in oxygen . the percentage yield of carbon dioxide is 82.5percent . what is the actual yield gram? why?​

Answer 1

To find the actual yield of carbon dioxide, we need to calculate the theoretical yield using the stoichiometry of the balanced chemical equation. From there, we can use the percentage yield to determine the actual yield. In this case, the actual yield of carbon dioxide is 134.33 grams.

Step-by-step explanation:

In order to find the actual yield of carbon dioxide, we first need to calculate the theoretical yield, which is the amount of carbon dioxide that would be produced if the reaction went to completion. From the balanced chemical equation, we can see that for every 2 molecules of octane burned, 16 molecules of carbon dioxide are produced.

Therefore, we can calculate the molar mass of octane (114.22 g/mol) and use stoichiometry to find the theoretical yield of carbon dioxide:

1. Convert the mass of octane (52.7 g) to moles: 52.7 g / 114.22 g/mol = 0.461 mol
2. Use the mole ratio from the balanced equation to find the moles of carbon dioxide: 0.461 mol octane * (16 mol CO₂ / 2 mol octane) = 3.69 mol CO₂
3. Convert the moles of carbon dioxide to grams using the molar mass of CO₂ (44.01 g/mol): 3.69 mol CO₂ * 44.01 g/mol = 162.45 g CO₂

Now that we have the theoretical yield of carbon dioxide, we can use the percentage yield (82.5%) to calculate the actual yield:

1. Actual yield = Percentage yield * Theoretical yield = 82.5% * 162.45 g = 134.33 g

Therefore, the actual yield of carbon dioxide is 134.33 grams.