To find the actual yield in grams, we need to first calculate the theoretical yield of CO2 using stoichiometry.
The balanced equation for the combustion of octane is:
2C8H18 + 25O2 → 16CO2 + 18H2O
From the balanced equation, we can see that the molar ratio between octane and CO2 is 2:16. This means that for every 2 moles of octane, 16 moles of CO2 are produced.
First, we need to convert grams of octane to moles. The molar mass of octane (C8H18) is (12.01 g/mol x 8) + (1.01 g/mol x 18) = 114.31 g/mol.
52.7 g of octane = (52.7 g) / (114.31 g/mol) = 0.461 mol octane
Now, using the mole ratio, we can calculate the theoretical yield of CO2.
0.461 mol octane x (16 mol CO2 / 2 mol octane) = 3.688 mol CO2
To find the actual yield of CO2 in grams, we multiply the moles of CO2 by the molar mass of CO2, which is 44.01 g/mol.
Actual yield = 3.688 mol CO2 x (44.01 g/mol) = 162.36 g
Therefore, the actual yield of CO2 in grams is 162.36 g.
In the reaction 2C8H18+25O2=16Co2+18H2 when 52.7g of octane(C8H18) burns in oxygen the persentage yield of Co2 is 82.5% what is the actual yield in gram?
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