When 52.7 g of octane (c8 h18) burns in oxygen, the percentage yield of carbon dioxide is 82.5%. what is the actual yield in grams?
When 52.7g of octane burns in oxygen,the percentage yield of carbon dioxide is 82.5%.what is the actual yield in grams? In the reaction 2c8h18+25o2=16co2+18h2o
2C8H18 + 25O2 ==> 16CO2 + 18H2O
mols C8H18 = grams/molar mass = approx 0.5 but that's just an estimate as are all the numbers that follow.
mols CO2 produced = approx 0.5 x 16/2 = approx 4.
approx g CO2 = mols cO2 x molar mass CO2 = ? This is the theoretical yield (TY)in grams. The actual yield (AY) is what you want.
%yield = [(AY)/(TY)]*100
You know % yield and TY, solve for AY. Post your work if you get stuck.
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43.4775
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2C8h18+2so2------>16co2+18H2O
52.7g=x
228g 764g
X=162.7g theoretical yield
Ac= 82.5×182.7g over 100
Actual yield =134.2g