Asked by Chera

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?
Answered by DrBob222
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.
Answered by Robel
Kalayu
Answered by Afnan
73.4g CO2
Answered by Beamlak
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
Answered by Hayat
Maths,chemistry,physics
Answered by netsanet
43.4775

Answered by Yeabsra
To develop my study skill
Answered by Yeabsra
I don't now this question 🙋
Answered by Mazinder
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
Answered by Nahom
135.25
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