Question

Consider the following balanced equation for the following reaction:

15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if the percent yield of CO2(g) is 83.0% and the theoretical yield of CO2(g) is 1.30 moles.

Answer 1

Answer: The amount of carbon dioxide formed in the reaction is 47.48 grams

Step-by-step explanation:

For the given chemical equation:

`15O_2(g)+2C_6H_5COOH(aq.)\rightarrow 14CO_2(g)+6H_2O(l)`

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 = 83 %

Theoretical yield of carbon dioxide = 1.30 moles

Putting values in above equation, we get:

`83=\frac{\text{Actual yield of carbon dioxide}}{1.30moles}* 100\\\\\text{Actual yield of carbon dioxide}=(1.30* 83)/(100)=1.079moles`

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.079 moles

Putting values in above equation, we get:

`1.079mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.079mol* 44g/mol)=47.48g`

Hence, the amount of carbon dioxide formed in the reaction is 47.48 grams