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

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

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

Step-by-step 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}=(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 =
`(2)/(2)* 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* 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}}* 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}=(6.70g)/(9.28g)* 100\\\\\% \text{yield of carbon dioxide}=72.2\%`

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