QQuestionChemistry
QuestionChemistry
What is the molar mass of butane, C^4H^10 ? I got 58.12 g/mol
Calculate the mass of water produced when 8.80 g of butane reacts with excess oxygen.
Calculate the mass of butane needed to produce 76.2 g of carbon dioxide.
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Answer
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Step 1: Find the molar mass of butane (C^4H^10).
\text{Molar mass of C4H10} = 4 \times 12.01~\text{g/mol} + 10 \times 1.01~\text{g/mol} = 58.12~\text{g/mol}
The molar mass of a compound is the sum of the molar masses of its constituent elements. For butane, we have: The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol. So,
Final Answer
Step 2: Calculate the mass of water (H^2O) produced when 8.80 g of butane reacts with excess oxygen. The balanced chemical equation for the complete combustion of butane is: \text{C^4H^10} + \frac{13}{2}~\text{O^2} \longrightarrow 4~\text{CO^2} + 5~\text{H^2O} From the balanced equation, we can see that for every mole of butane that reacts, 5 moles of water are produced. Using the molar mass calculated in Step 1, we can find the number of moles of butane in the given sample: 8.80~\text{g}~\text{C^4H^10} \times \frac{1~\text{mol}~\text{C^4H^10}}{58.12~\text{g}~\text{C^4H^10}} \approx 0.1515~\text{mol}~\text{C^4H^10} Now, we can calculate the mass of water produced: 0.1515~\text{mol}~\text{C^4H^10} \times \frac{5~\text{mol}~\text{H^2O}}{1~\text{mol}~\text{C^4H^10}} \times \frac{18.02~\text{g}~\text{H^2O}}{1~\text{mol}~\text{H^2O}} \approx 13.73~\text{g}~\text{H^2O} Step 3: Calculate the mass of butane needed to produce 76.2 g of carbon dioxide (CO2). The balanced chemical equation for the complete combustion of butane is the same as in Step 2. From the balanced equation, we can see that for every 4 moles of butane that react, 4 moles of carbon dioxide are produced. We can set up a proportion to find the mass of butane needed: \frac{76.2~\text{g}~\text{CO^2}}{4~\text{mol}~\text{CO^2}} = \frac{58.12~\text{g}~\text{C^4H^10}}{1~\text{mol}~\text{C^4H^10}} Solving for the mass of butane, we get: 58.12~\text{g}~\text{C^4H^10} \times \frac{76.2~\text{g}~\text{CO^2}}{4 \times 58.12~\text{g}~\text{C^4H^10}} \approx 50.95~\text{g}~\text{C^4H^10}
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