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

Please show your working using an ICE table.

Answer 1

2 C₈H₁₈    +     25 O₂ --------------------> 16 CO₂ + 18 H₂O

228.46 g          800 g                                 704 g         324 g

   2                   25                                        16              18

4                     8                                           ??

here limting reagent is O2.

25 mol O2   ----------------> 16 mol CO2

8 mol O2    ----------------->   ??

moles of CO2 formed = 8 x 16 / 25 = 5.12 mol

theoretical yield = 5.12 x 44 = 225.28 g

percent yield = actual / theoretical ) x 100

                     = 28.16 / 225.28 ) x 100

percent yield = 12.5 %