Oxygen gas can be generated by heating KClO3 in the presence of a catalyst.
KClO3 KCl + O2 (unbalanced)
What volume of O2 gas will be generated at T = 35°C and P = 764.0 torr from 1.51 g of KClO3?
I got .414 L and used 22.4 L for moles of O2
What's going on??

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  1. 22.4 L/mol is the volume at STP and you don't have STP conditions.
    Balance the equation, find n O2, then use PV = nRT and substitute the conditions of P and T listed (remember T must be in kelvin) and solve for V in L.

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  2. I would convert torr to atm (torr/760) and use 0.08206 for R.

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  3. Ok thanks for your help!

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  4. You're welcome.

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