Questions: Calculate the mass of butane needed to produce 73.0 g of carbon dioxide. Express your answer to three significant figures and include the appropriate units.

Solution Steps

The combustion of butane (\(\text{C}_4\text{H}_{10}\)) in oxygen produces carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). The balanced chemical equation for this reaction is:

\[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \]

Calculate the molar mass of butane and carbon dioxide:

• Molar mass of \(\text{C}_4\text{H}_{10}\): \[ (4 \times 12.01) + (10 \times 1.008) = 58.12 \, \text{g/mol} \]

• Molar mass of \(\text{CO}_2\): \[ (1 \times 12.01) + (2 \times 16.00) = 44.01 \, \text{g/mol} \]

From the balanced equation, 8 moles of \(\text{CO}_2\) are produced from 2 moles of \(\text{C}_4\text{H}_{10}\). First, calculate the moles of \(\text{CO}_2\) produced:

\[ \text{Moles of } \text{CO}_2 = \frac{73.0 \, \text{g}}{44.01 \, \text{g/mol}} = 1.659 \, \text{mol} \]

Using the stoichiometry of the reaction:

\[ \text{Moles of } \text{C}_4\text{H}_{10} = \frac{1.659 \, \text{mol} \times 2}{8} = 0.4148 \, \text{mol} \]

Finally, convert the moles of butane to grams:

\[ \text{Mass of } \text{C}_4\text{H}_{10} = 0.4148 \, \text{mol} \times 58.12 \, \text{g/mol} = 24.11 \, \text{g} \]

Final Answer