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 actual yield of the carbon dioxide is 47.48 grams

Explanation:

For the given balanced equation:

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

To calculate the mass for given number of moles, we use the equation:

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

Theoretical moles of carbon dioxide = 1.30 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

$1.30mol=\frac{\text{Theoretical yield of carbon dioxide}}{44g/mol}\\\\\text{Theoretical yield of carbon dioxide}=(1.30mol\times 44g/mol)=57.2g$

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

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Theoretical yield of carbon dioxide = 57.2 g

Percentage yield of carbon dioxide = 83.0 %

Putting values in above equation, we get:

$83=\frac{\text{Actual yield of carbon dioxide}}{57.2}\times 100\\\\\text{Actual yield of carbon dioxide}=\frac{57.2\times 83}{100}=47.48g$

Hence, the actual yield of the carbon dioxide is 47.48 grams