Question

give the percent yield when 39.92 g of Co2 are formed from the reaction of 2.43 moles of C8H18 with 4.86 moles of O2

Answer 1

Consider a reaction 2 C₈H₁₈ + 25 O ₂16 CO₂ + 18 H₂O

From reaction , 2 mole octane combines with 25 mole oxygen to give 16 moles of carbon dioxide.

Hence stoichiometric ratio of reactants is C₈H₁₈ : O ₂ = 2 : 25 = 1 : 12.5

Provided molar ratio of reactants is  C₈H₁₈ : O ₂ = 2.43 : 4.86 = 1: 2

Comparing provided molar ratio with stoichiometric ratio of reactants it is clear that O ₂ is limiting reactant , hence yield of product will depend upon amount of O ₂.

From reaction we can write,

25 mole O ₂ 16 mole CO₂

4.86 mole O ₂ 16 x 4.86 / 25 mole CO₂

   3.11 mole CO₂

3.11 mole CO₂ = 3.11 mol x ( 44.01 g / mol ) = 136.9 g CO₂

Theoretical yield of product = 136.9 g

Practical yield of product = 39.92 g  

% practical yield = [Practical yield of product / Theoretical yield of product ] x 100

= ( 39.92 g / 136.9 g ) x 100

= 29.2 %