Calculating O2 and Bi2S3 for Volume Law of Combining Volumes

[htk]

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

 

[Borek]

Approach to the first part seems correct (even if there is no SO₂ between products ).

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

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[Blueshift5]

I would first check the stoichiometry of the reaction to ensure that the coefficients are correct. Based on the given reaction, it seems like there is a typo and the product should be 2Bi2O3 + 6SO2, not 6O2. Assuming that is the case, your answer for the amount of O2 needed is correct.

To calculate the amount of Bi2S3 needed, we can use the volume law of combining volumes, which states that the volumes of gases in a chemical reaction are in a simple ratio to the number of moles. In this case, we can use the ideal gas law to calculate the number of moles of O2 needed to produce 3.87 L of SO2 at 1.00 atm and 174 C:

n = PV/RT = (1.00 atm)(3.87 L)/(0.0821 L*atm/mol*K)(447 K) = 0.1165 mol O2

Since the stoichiometric ratio between O2 and Bi2S3 is 9:2, we can calculate the number of moles of Bi2S3 needed:

n(Bi2S3) = (2/9)(0.1165 mol O2) = 0.0259 mol Bi2S3

Finally, we can use the molar mass of Bi2S3 to calculate the mass needed:

m(Bi2S3) = 0.0259 mol x 514.155 g/mol = 13.34 g Bi2S3

Therefore, the correct answer for the amount of Bi2S3 needed is 13.34 g, not 663.26 g. I hope this helps clarify any confusion and reinforces the importance of checking the stoichiometry in a chemical reaction.