Question

Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 8.000 moles of O2.
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

Answer 1

From the balanced equation we can say that

2 mole of C8H8 requires 25 mole of O2 so

4.00 mole of C8H8 will require

= 4.00 mole of C8H8 *( 25 mole of O2 / 2 mole of C8H8)

= 50.00 mole of O2

But we have 8.000 mole of O2 which is in short so O2 is limiting reactant

From the balanced equation we can say that

25 mole of O2 produces 16 mole of CO2 so

8.000 mole of O2 will produce

= 8.000 mole of O2 *(16 mole of CO2 / 25 mole of O2)

= 5.12 mole of CO2

mass of 1 mole of CO2 = 44.01 g so

the mass of 5.12 mole of CO2 = 225 g

Therefore, theoretical yield of CO2 = 225 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (28.16 / 225)*100 = 12.5

Therefore, percent yield = 12.5 %