Consider the balanced equation for the following reaction:
7O2(g) + 2C2H6(g) → 4CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.

The amount of carbon dioxide formed in the reaction is 0.566 grams

Explanation:

To calculate the number of moles, we use the equation:

      .......(1)

Mass of oxygen gas = 8.00 grams

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

As, ethane is present in excess. It is considered as an excess reagent.

So, oxygen gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in above equation, we get:

To calculate the actual yield of carbon dioxide, we use the equation:

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.29 g

Putting values in above equation, we get:

Hence, the amount of carbon dioxide formed in the reaction is 0.566 grams

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