A115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2kclo3 mc019-1.jpg 2kci + 3o2 what is the percent yield of oxygen in this chemical reaction? use mc019-2.jpg.

The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%