Question #1a855

You can't actually solve this problem without knowing the conditions for pressure and temperature at which the oxygen gas is collected.

The idea here is that you need to use the given volume to determine how many moles of oxygen gas were produced by the reaction.

Since no mention of pressure and temperature was made, I"ll assume that you're working at STP - Standard Temperature and Pressure.

STP conditions imply a pressure of #"100 kPa"# and a temperatue of #0^@"C"#. Under these conditions, one mole of any ideal gas occupies exactly #"22.7 L"# - this is known as the molar volume of a gas at STP.

In your case, #"108 cm"^3# of oxygen gas would contain

#108color(red)(cancel(color(black)("cm"^3))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 mole"/(22.7color(red)(cancel(color(black)("L")))) = "0.0047577 moles O"_2#

Now take a look at the balnced chemical equation

#color(red)(2)"KClO"_text(3(s]) -> 2"KCl"_text((s]) + color(blue)(3)"O"_text(2(g]) uarr#

Notice that you have a #color(red)(2):color(blue)(3)# mole ratio between potassium chlorate and oxygen gas.

This means that if the reaction produced #0.0047577# moles of oxygen gas, it consumed

#0.0047577color(red)(cancel(color(black)("moles O"_2))) * (color(red)(2)" moles KClO"_3)/(color(blue)(3)color(red)(cancel(color(black)("KClO"_3)))) = "0.003172 moles KClO"_3#

To find how many grams of potassium chlorate would contain this many moles, use the compound's molar mass

#0.003172color(red)(cancel(color(black)("moles"))) * "122.55 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.389 g KClO"_3)#