Rate of Production with Given Information

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SUMMARY

The discussion focuses on calculating the rate of oxygen (O2) release from the decomposition of potassium chlorate (KClO3) under standard ambient temperature and pressure (SATP). The decomposition reaction is represented as 2 KClO3 (s) + heat --> 2 KCl (s) + 3 O2 (g). The participant successfully converted 0.20 g of KClO3 to 0.001632 mol and determined the rate of decomposition to be 0.000583 mol/s. The next step involves applying the ideal gas law (n = PV/RT) to find the rate of O2 production.

PREREQUISITES
  • Understanding of stoichiometry and molar conversions
  • Familiarity with the ideal gas law (n = PV/RT)
  • Knowledge of standard ambient temperature and pressure (SATP) conditions
  • Basic principles of chemical kinetics
NEXT STEPS
  • Apply the ideal gas law to calculate the volume of O2 produced at SATP
  • Investigate the relationship between molar ratios in chemical reactions
  • Explore the concept of reaction rates and their measurement
  • Review the principles of gas laws and their applications in chemical reactions
USEFUL FOR

Chemistry students, educators, and anyone studying chemical kinetics and gas laws will benefit from this discussion, particularly those focusing on reaction rates and stoichiometric calculations.

Complex_
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Homework Statement



2 KClO3 (s) + heat --> 2 KCl (s) + 3 O2 (g)

If 0.20 g of KClO3 (s) decomposes in 2.8 s, at what rate is oxygen O2 (g) released over this time at SATP?

Homework Equations



n = PV/RT

The Attempt at a Solution



Converted KClO3 to mol.

.20 x 1 / 122.5495g = 0.001632 mol KCl

Find rate of decomposition :

.001632 mol in 2.8 sec --> 2 mol in 3431.37 sec

2/3431.37 = .000583 mol/s

The rate of decomposition of KClO3 is .000583 mol/s

I'm stuck at this part. I know I need to use some sort of n = PV/RT since it tells me that it is at SATP. Can I have some hints or help?
 
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Complex_ said:
0.001632 mol KCl
Check.
Complex_ said:
.000583 mol/s
Check.
Complex_ said:
at what rate is oxygen O2 (g)
What's next?
 

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