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What volume of oxygen gas can be collected at 0.565 atm pressure and 48.0◦C when 42.9g of KClO3 decompose by heating, according...Question
What volume of oxygen gas can be collected
at 0.565 atm pressure and 48.0◦C when 42.9
g of KClO3 decompose by heating, according
to the following equation?
2 KClO3(s) ∆−−−−→2KCl(s) + 3O2(g)
MnO2
at 0.565 atm pressure and 48.0◦C when 42.9
g of KClO3 decompose by heating, according
to the following equation?
2 KClO3(s) ∆−−−−→2KCl(s) + 3O2(g)
MnO2
Answers
DrBob222
mols KClO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2.
Now use PV = nRT at the conditions listed to calculate volume in L O2.
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2.
Now use PV = nRT at the conditions listed to calculate volume in L O2.
Anon
I can get to the PV=nRT part. What number would you put for R?
DrBob222
You should have told me you can get to that and we could have cut out half of the work. If you use atm for P, then R is 0.08206 L*atm/mol*K.
Anon
What I was saying in the response was that with your help I got to that part, but I was still confused on the formula. I got it now. Thank you.
DrBob222
You're welcome. I'm sorry I misunderstood.