Question

Combustion of 8.9541 grams of C4H10 will yield (Example answer: 2.345) Do not enter units. grams of CO2. (Assume 100% yield)

Answer 1

Answer: The mass of carbon dioxide produced is 27.1 grams.

Explanation:

Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

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 butane = 8.9541 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of butane}=\frac{8.9541g}{58.12g/mol}=0.154mol$

The chemical equation for the combustion of butane follows:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 0.154 moles of butane will produce = $\frac{8}{2}\times 0.154=0.616mol$ of carbon dioxide

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

Moles of carbon dioxide = 0.616 moles

Mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

$0.616mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=27.1g$

Hence, the mass of carbon dioxide produced is 27.1 grams.