Question

Consider the following balanced equation for the following reaction:

Answer 1

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

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:

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

Percentage yield of carbon dioxide = 83 %

Theoretical yield of carbon dioxide = 1.30 moles

Putting values in above equation, we get:

$83=\frac{\text{Actual yield of carbon dioxide}}{1.30moles}\times 100\\\\\text{Actual yield of carbon dioxide}=\frac{1.30\times 83}{100}=1.079moles$

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:

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

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