Question

4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen.

A) Write a balanced equation. (ethylene plus oxygen yields carbon dioxide and water)
i) 2C2H4 + 3O2 → 2CO2 + 2H2O

B) What is the limiting reactant? You must show your work.
i) Work to be shown.

C) If the actual yield of Carbon dioxide is 2.50 g, what is the percent yield for the reaction? You must show your work.
i) Work to be shown.

Answer 1

The balanced chemical equation for the combustion of ethylene is 2C2H4 + 7O2 → 4CO2 + 4H2O. Oxygen is the limiting reactant and the percent yield of carbon dioxide in the reaction is 97.28%.

Step-by-step explanation:

Combustion of Ethylene

First, the balanced chemical equation for the combustion of ethylene (C2H4) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is:

2C2H4 + 7O2 → 4CO2 + 4H2O

To find the limiting reactant, we calculate the moles of each reactant:

Molar mass of C2H4 = 28 g/mol
Moles of C2H4 = 4.95 g / 28 g/mol = 0.177 moles

Molar mass of O2 = 32 g/mol
Moles of O2 = 3.25 g / 32 g/mol = 0.102 moles

Using the stoichiometry of the balanced equation, for every 2 moles of C2H4, you need 7 moles of O2. Therefore, for 0.177 moles of C2H4, you would need 0.177 moles x (7 moles O2 / 2 moles C2H4) = 0.619 moles of O2. Since we only have 0.102 moles of O2, O2 is the limiting reactant.

To calculate the percent yield, we first need to find the theoretical yield of CO2. From the balanced equation, 2 moles of C2H4 produce 4 moles of CO2. Using the limiting reactant (O2):

1. Moles of CO2 produced per mole of O2 = 4 moles CO2 / 7 moles O2
2. Theoretical moles of CO2 = 0.102 moles O2 x (4 moles CO2 / 7 moles O2) = 0.0583 moles CO2
3. Theoretical yield (g) of CO2 = 0.0583 moles x 44 g/mol (molar mass of CO2) = 2.57 g
4. Percent yield = (actual yield / theoretical yield) x 100 = (2.50 g / 2.57 g) x 100 = 97.28%

The percent yield of the carbon dioxide in the reaction is 97.28%.