We're asked to find the mass, in #"g"#, of #"KClO"_3# that decomposes to give a certain amount of #"O"_2#.
Let's use the ideal gas equation to find the moles of #"O"_2# that form.
#P = 747cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 0.983# #"atm"#
#V= 797cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = 0.797# #"L"#
#T = 128^"o""C" + 273 = 401# #"K"#
Plugging in known values, we have
#n = (PV)/(RT) = ((0.983cancel("atm"))(0.797cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(401cancel("K"))) = 0.0238# #"mol O"_2#
Now, we'll use the coefficients of the chemical equation to find the relative number of moles of #"KClO"_3# that reacted:
#0.0238cancel("mol O"_2)((2color(white)(l)"mol KClO"_3)/(3cancel("mol O"_2))) = 0.0159# #"mol KClO"_3#
Finally, we'll use the molar mass of potassium chlorate (#122.55# #"g/mol"#) to find the number of grams that reacted:
#0.0159cancel("mol KClO"_3)((122.55color(white)(l)"g KClO"_3)/(1cancel("mol O"_2))) = color(blue)(1.95# #color(blue)("g KClO"_3#
