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.
