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Andre45 [30]
3 years ago
9

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

Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

<u>Answer:</u> The mass of carbon dioxide produced is 27.1 grams.

<u>Explanation:</u>

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.

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