Atmospheric Stoichiochemistry: what mass of CO2 is produced?

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SUMMARY

The discussion centers on calculating the mass of carbon dioxide produced from burning 15.5 L of acetylene gas at STP, resulting in 60.9 g of CO2. The ideal gas law (PV = nRT) was applied to determine that 0.692 moles of acetylene yield 1.384 moles of CO2 based on the reaction 2C2H2 + 5O2 → 2H2O + 4CO2. Despite achieving the correct mass, the relevance of the conditions of 1500 °C and 92.5 kPa was questioned, with some participants suggesting they impact the volume of gas produced but not the mass calculation.

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I have two problems actually:

Problem 1

Homework Statement



What mass of carbon dioxide can be produced at a temperature of 1500 *C and an atmospheric pressure of 92.5 kPa, if 15.5 L of acetylene gas is burned at STP?

Homework Equations



PV = nRT
P1V1/T1 = P2V2/T2

The Attempt at a Solution



First, I found the amount of moles of acetylene gas, using the ideal gas law, PV = nRT
nacetylene = 0.692 mol

Then, I used a mole ratio according to the equation to find the moles of carbon dioxide gas:

2C2H2 + 5O2 → 2H2O + 4CO2

And 1.384 mol of carbon dioxide gas should be produced, multiplied by its molar mass of 32, I got 60.9 g.

What confuses me is the 1500 *C and 92.5 kPa. The question was out of four marks and my solution seems worth that many marks. However, when I asked about it, my chemistry teacher sternly implied that the 1500 C and 92.5 kPa is relevant. This is my alternative solution, that gives the same answer, requires more work but does account for both the above values:

Used mole ratio to go from 15.5 L acetylene to 31 L carbon dioxide gas. Converted from STP (31 L) to the given conditions, which satisfy the ideal conditions for an ideal gas that I found online (which was not in the course) that says that gases are "more" ideal at high temperature and low pressure: 92.5 kPa < STP, 1500 C > STP.

I now had 220.5 L volume at the temperature and pressure given and used the ideal gas law to solve, and got the same answer:

92.5 kPa * 220.5 L/ 8.31 J * mol-1 * K-1 * 1773 K = 1.38 mol.

I got zero on the question.

Thanks in advance for ANY AND ALL input! :)

I think the second one is slightly more appropriate, but both directions give the same answer, and considering that I barely had time to finish the test, the first approach seemed more practical.
 
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Either the teacher meant something else than the question asks, or you are quoting it incorrectly - but the way it is worded neither 1500°C nor 92.5 kPa matters. They would matter for the volume of gas produced, but not for the mass.

15.5 L of acetylene at STP is 0.692 moles, and it produces 60.9 g of CO2, these are perfectly correct numbers.



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