Question

Hello! Please help.

4.38 grams of CO2 was obtained when 1.93 grams of a mixture of ethylene (C2H4) and 5.92 grams of O2 is ignited according to the following equation. Calculate the percentage yield of CO2 if the theoretical yield is found to be 5.43 grams of CO2?C2H4 + 3O2 ---> 2CO2 + 2H2O

Answer 1

Answer:

C2H4 + 3O2 ==> 2CO2 + 2H2O

Find the limiting reactant:

moles ethylene = 1.93 g x 1 mole/28.05 g = 0.06881 moles C2H4 = 0.1376 moles CO2 produced

moles O2 = 5.92 g x 1 mole/32 g = 0.185 moles O2 = 0.1233 moles CO2 produced

O2 is limiting

mass CO2 produced = 0.1233 moles CO2 x 44 g/mole = 5.43 g CO2 produced

Answer 2

Answer:

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Explanation:

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