How many grams are in one mole of CH3CO2H ?
1 answer:
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Percent yield is 23.11 % when 162.8 g of CO2 are formed from the reaction of excess amount of C8H18 and with 218.0 grams of O2.
Explanation:
Balanced equation for the chemical reaction:
2C8H18 + 25O2 → 16CO2 + 18H2O
data given:
CO2 formed (actual yield) = 162.8 grams
mass of oxygen = 218 grams
16 moles of CO2 formed when 5 moles of oxygen reacted
3.6 moles of CO2 formed when 6.8 moles of oxygen reacted.
In the reaction 16 moles of CO2 will have 44.01 x 16
theoretical yield of CO2 = 704.16 grams
percent yield = x100
putting the values in the above equation
percent yield = x 100
= 0.23 x 100
= 23.11 %
Percent yield is 23.11 %.