The decomposition of potassium chlorate, #"KClO"_3#, yields the salt potassium chloride, #"KCl"#, and oxygen, #"O"_2"#. The balanced chemical equation is:
#"2KClO"_3("s")"##rarr##"2KCl(s)"# + #"3O"_2"#
We need the molar masses of #"KClO"_3# and #"O"_2#.
Molar mass of #"KClO"_3# = #"122.55 g/mol"#.
Molar mass of #"O"_2# = #"31.998 g/mol"#.
We will use the molar masses and mole ratio from the balanced equation to determine the amount of #"O"_2"# that can be formed from #"3.76 g KClO"_3"#
Step 1. Convert #"3.76 g of KClO"_3# to moles.
#3.76 color(red)cancel(color(black)("g KClO"_3))xx (1"mol KClO3")/(122.55color(red)cancel(color(black)("g KClO3")))= "0.030681 mol KClO"_3"#
I am keeping a couple of extra digits to reduce rounding errors. I will round the final answer to three significant figures.
Step 2. Use the mole ratio between #"KClO"_3# and #"O"_2# to determine the number of moles of #"O"_2# can be produced from #"0.030681 mol KClO"_3#.
The mole ratio between #"KClO"_3# and #"O"_2# is #"2 mol KClO"_3# : #"3O"_2#.
Multiply the calculated moles of #"KClO"_3"# times the mole ratio with #"O"_2# on top.
#0.030681 color(red)cancel(color(black)("mol KClO"_3))xx(3 "mol O2")/(2 color(red)cancel(color(black)("mol KClO"_3)))=" 0.046022 mol O2"#
Step 3. Convert moles of #"O"_2# to grams by multiplying times its molar mass.
#0.046022 color(red)cancel(color(black)("mol O"_2))xx(31.998 "g O2")/(1 color(red)cancel(color(black)("mol O"_2)))= "1.47 g O"_2# (rounded to 3 sig figs)
The decomposition of #"3.76 g KClO"_3"# produces #"1.47 g O"_2"#.
