Question

For the following reaction, 5.91 grams of carbon monoxide are mixed with excess oxygen gas . The reaction yields 6.70 grams of carbon dioxide . carbon monoxide ( g ) + oxygen ( g ) carbon dioxide ( g )
a. What is the theoretical yield of carbon dioxide ? grams
b. What is the percent yield for this reaction ? %

Answer 1

Answer:

For a: The theoretical yield of carbon dioxide is 9.28 grams.

For b: The percent yield of the reaction is 72.2 %.

Explanation:

• For a:

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

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

Given mass of carbon monoxide = 5.91 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{5.91g}{28g/mol}=0.211mol$

The chemical equation for the reaction of carbon monoxide and oxygen gas follows:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

By Stoichiometry of the reaction:

2 moles of carbon monoxide produces 2 moles of carbon dioxide

So, 0.211 moles of carbon monoxide will produce = $\frac{2}{2}\times 0.211=0.211mol$ of carbon dioxide

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

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.211 mol

Putting values in equation 1, we get:

$0.211mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.211mol\times 44g/mol)=9.28g$

Hence, the theoretical yield of carbon dioxide is 9.28 grams.

• For b:

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

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

Experimental yield of carbon dioxide = 6.70 g

Theoretical yield of carbon dioxide = 9.28 g

Putting values in above equation, we get:

$\%\text{ yield of carbon dioxide}=\frac{6.70g}{9.28g}\times 100\\\\\% \text{yield of carbon dioxide}=72.2\%$

Hence, the percent yield of the reaction is 72.2 %.