Give the percent yield when 28.16 g of CO2 are formed from the reaction of 5.000 moles of C8H18 with 10.00 moles of O2.
2 C8H18 + 25O2 → 16 CO2 + 18H2O
First, we need to determine the limiting reactant in the reaction. To do this, we will calculate the amount of CO2 that can be produced from both reactants and see which one produces less CO2.
From the balanced chemical equation, we see that 2 moles of C8H18 produces 16 moles of CO2.
5.000 moles of C8H18 × (16 moles CO2 / 2 moles C8H18) = 40.00 moles of CO2
From the balanced chemical equation, we see that 10 moles of O2 produces 16 moles of CO2.
10.00 moles of O2 × (16 moles CO2 / 25 moles O2) = 6.400 moles of CO2
Therefore, O2 is the limiting reactant.
Now, we can calculate the theoretical yield of CO2 from the reaction of 10.00 moles of O2.
10.00 moles of O2 × (16 moles CO2 / 25 moles O2) × (44.01 g CO2 / 1 mole CO2) = 28.16 g CO2
Given: actual yield = 28.16 g CO2
Theoretical yield = 28.16 g CO2
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (28.16 g / 28.16 g) x 100 = 100%
Therefore, the percent yield of CO2 is 100%.