If 52 grams of C2H4 is burned in the presence of 144 grams of oxygen, what is the % yield of the reaction if 90 grams of carbon dioxide is produced in the reaction?

Question

For the following reaction:

C2H4 + 3O2 —> 2H2O + 2CO2

If 52 grams of C2H4 is burned in the presence of 144 grams of oxygen, what is the % yield of the reaction if 90 grams of carbon dioxide is produced in the reaction? Remember, you must calculate your answer based on limiting reagent.

Answer 1

As I was describing in your previous question, you can find the limiting reactant by dividing moles of reactant by the coefficient in the balanced equation.

For C2H4: 52 g C2H4 x 1 mol/28 g = 1.86 moles (divided by 1 = 1.86)

For O2: 144 g O2 x 1 mol/32 g = 4.5 moles (divided by 3 = 1.5). O2 is limiting b/c 1.5 is less than 1.86.

Use the 4.5 moles O2 to find theoretical yield of CO2 and then find % yield as actual yield/theoretical (x100)

Answer 2

One approach is to use the typical 3-step stoichiometry technique to get the grams of product. Do the calculation for the given grams of C₂H₄ with excess oxygen, then do the same calculation for the g of O₂ with excess ethylene. The one that produces the least CO₂ is the limiting reactant and the amount of CO₂ produced ideally (theoretical yield) will be that lower amount.

The percent yield is defined as (actual yield (given in problem) / theoretical yield) x 100

I will show the technique for g CO₂ from g C₂H₄:

52 g C₂H₄ x (1mole/28.06g C₂H₄) x (2 mole CO₂/ 1 mole C₂H₄) x (44.01 g/mole CO₂) = 163.12 g CO₂

You have to do the same thing for g O2 to g CO2.

Good luck!