We're asked to find the number of moles of #"KClO"_3# that need to decompose to yield #67.2# #"L O"_2# (assuming #100%# yield).
The decomposition reaction of potassium chlorate is
#2"KClO"_3(s) rarr 2"KCl"(s) + 3"O"_2(g)#
At s.t.p., one mole of any (ideal) gas occupied a volume of #22.41# #"L"#, so we can use this to convert from liters of oxygen to moles of oxygen:
#67.2cancel("L O"_2)((1color(white)(l)"mol O"_2)/(22.41cancel("L O"_2))) = color(red)(3.00# #color(red)("mol O"_3#
Now, we'll use the coefficients of the chemical equation to calculate the relative number of moles of #"KClO"_3# that need to decompose:
#color(red)(3.00)cancel(color(red)("mol O"_2))((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = color(blue)(2# #color(blue)("mol KClO"_3#
Thus, option (B) is correct.
