Question

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%.

Answer 1

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .......(1)

Mass of oxygen gas = 8.00 grams

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(8g)/(32g/mol)=0.25mol`

For the given chemical equation:

`7O_2(g)+2C_5H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)`

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 =
`(4)/(7)* 0.25=0.143moles` 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:

`0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol* 44g/mol)=6.29g`

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

`\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}* 100`

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.29 g

Putting values in above equation, we get:

`90=\frac{\text{Actual yield of carbon dioxide}}{6.29g}* 100\\\\\text{Actual yield of carbon dioxide}=(6.29* 90)/(100)=0.566g`

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