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?
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.