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Oxygen gas can be generated by heating KClO3 in the presence of a catalyst. KClO3 ----> KCl + O2 (unbalanced) What volume (in L...Asked by Anonymous
Oxygen gas can be generated by heating KClO3 in the presence of a catalyst.
KClO3 → KCl + O2 (unbalanced)
What volume (in L) of O2 gas will be generated at T = 25 oC and p = 1.017 bar from 3.44 g KClO3?
KClO3 → KCl + O2 (unbalanced)
What volume (in L) of O2 gas will be generated at T = 25 oC and p = 1.017 bar from 3.44 g KClO3?
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Answered by
oobleck
how many moles in 3.44g KClO3?
Balance the equation to find out how many moles of O2 you will get.
Since each mole occupies 22.4L at STP, if you get M moles, then you want V such that
1.017V/(273+25) = 1*22.4M/273
Balance the equation to find out how many moles of O2 you will get.
Since each mole occupies 22.4L at STP, if you get M moles, then you want V such that
1.017V/(273+25) = 1*22.4M/273
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