First, we must write the balanced equation for the decomposition of potassium chlorate:

#2KClO_3rarr2KCl+3O_2#

Note that the potassium chlorate and the oxygen are in the ratio #2:3#. That is to say, for every #2# moles of potassium chlorate which react, #3# moles of oxygen are produced.

The molar mass of potassium chlorate is #122.55"g/mol"#.

The formula for moles, #n#, is #n=m/M#, where #m# is the given mass and #M# the molar mass.

#n=283/122.55~~2.31"mol"# of potassium chlorate.

Remember that for every #2# moles of potassium chlorate which react, #3# moles of oxygen are produced.

So when #x# moles of oxygen are produced,

#2/3=2.31/x#

#2x=3*2.31#

#x~~3.47"mol"# of oxygen is produced.

You can rearrange the formula for moles to solve for #m#:

#n=m/M#

#m=nM#

The molar mass of oxygen is #32"g/mol"#, so we can input:

#m=32*3.47#

#m=110.88 \ "g"# of oxygen is produced.