Calculating O2 and Bi2S3 for Volume Law of Combining Volumes

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SUMMARY

The discussion focuses on calculating the volume of O2 and the mass of Bi2S3 required for the reaction 2Bi2S3 + 9O2 → 2Bi2O3 + 6SO2 at 1.00 atm and 174°C. The correct calculation for the volume of O2 needed to produce 3.87 L of SO2 is 5.805 L, derived using the stoichiometric ratio of 9 L O2 to 6 L SO2. To find the grams of Bi2S3 required, the calculation yields 663.26 g, based on the molar mass of Bi2S3 and the stoichiometric conversion from moles of O2 to moles of Bi2S3.

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htk
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Homework Statement


Consider the reaction 2Bi2S3 +9O2 ---------> 2Bi2O3 +6O2

The reaction is carried out and kept at 1.00 atm pressure and 174 C throughout. How many liters of O2 are needed to produce 3.87 L of SO2? How many grams of Bi2S3 are needed?

The Attempt at a Solution



my answer is 5.805 L O2 and 663.26 g Bi2S3. However, I don't know it is correct or incorrect. If it is wrong please explain to me why. Thank you!

3.87 LSO2 * 9LO2/ 6 L SO2 = 5.805 LO2

5.805 mol O2 * 2molBi2S3/ 9molO2= 1.29 mol Bi2S3 * 514.155 g Bi2S3/1mol Bi2S3= 663.26g Bi2S3
 
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Approach to the first part seems correct (even if there is no SO2 between products :wink: ).

However, 5.805 L of O2 is not 5.805 moles of O2.

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