Decomposition of potassium chlorate (kclo3) produces potassium chloride (kcl) and pure oxygen (o2). the balanced equation for the reaction is as follows.
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what volume of oxygen gas is released at stp if 10.0 g of potassium chlorate is decomposed? (the molar mass of kclo3 is 122.55 g/mol.)
0.914 l
1.83 l
2.74 l
3.66 l

1.    The balanced reaction would be:

2KClO3 --> 2KCl + 3O2

We are given the amount of potassium chlorate for the reaction. This will be the starting point of our calculation. 

10.0 g KClO3 ( 1 mol KClO3 / 122.55 g KClO3) (3 mol O2 / 2 mol KClO3) = 0.1224 mol O2

V = nRT/P =0.1224 mol O2 x 273 K x 0.08206 atm L/mol K / 1 atm
V=2.74 L