Question

What is the percent yield of CO2 if 42.24 g of CO2 are formed from the reaction of 4.00 moles of C8H18 with 8.00 moles of O2 according to the balanced equation below? 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

Answer 1

The reaction is 2C8H18+25O2----> 16CO2 +18 H2O

As per the reaction, molar ratio of reactants = 2: 25 =1:12.5

given molar ratio of reactants = 4: 8= 1: 2

Oxygen is the limiting reactant. since it is required in 12.5 for 1mole of C8H18 while supplied is 8 only. It controls the extent of reaction

25 moles oxygen consumes 2moles of C8H18

8 moles of oxygen (limiting reactant) consumes 8*2/25=0.64 mole of C8H18

25 moles oxygen prdices 16 moles of CO2.

8 moles oxygen produces 8*16/25=5.12 moles of CO2. CO2 prdoced= 42.24 gm moles of CO2 produced= mass/Molecualr weight= 42.24/44=0.96

Percent yiled= 100* actual yield/ theoretical yield = 100*0.96/5.12=18.75%