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Balance the following reaction first: \( \frac{\mathrm{C}_{3} \mathrm{H}_{3}}{}+\frac{\mathrm{O}_{2} \rightarrow}{} \) grams of water will be produced from 12.0 liters of oxygen?

Ask by Li Hardy. in the United States
Feb 25,2025

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Approximately 19.3 grams of water will be produced from 12.0 liters of oxygen.

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To balance the reaction \( \mathrm{C}_{3} \mathrm{H}_{3} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \), we first need to find the correct coefficients. The balanced equation is: \[ \mathrm{C}_{3} \mathrm{H}_{3} + 4 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_2 + 3 \mathrm{H}_2\mathrm{O} \] Now, using the ideal gas law, we know that at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, in 12.0 liters of \( \mathrm{O}_{2} \): \[ \text{Moles of } \mathrm{O}_{2} = \frac{12.0 \text{ L}}{22.4 \text{ L/mol}} \approx 0.536 \text{ moles} \] From the balanced equation, 4 moles of \( \mathrm{O}_{2} \) produce 3 moles of \( \mathrm{H}_2\mathrm{O} \). So: \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{3}{4} \times \text{Moles of } \mathrm{O}_{2} \] Calculating the moles of water produced: \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{3}{4} \times 0.536 \approx 0.402 \text{ moles} \] Finally, the mass of water produced can be calculated using the molar mass of water (approximately 18 g/mol): \[ \text{Mass of } \mathrm{H}_2\mathrm{O} = 0.402 \text{ moles} \times 18 \text{ g/mol} \approx 7.236 \text{ grams} \] So, approximately 7.24 grams of water will be produced from 12.0 liters of oxygen.

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