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 %.
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
All are the same elements but they differ in their atomic mass therefore this process will called Isotopes