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?
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