How many liters of oxygen are produced from the decomposition of 122g of potassium chlorate?
The first thing you ought to do is to convert 122 g into moles. 122 g KClO3 * 1 mol KClO3 / 122.55 g KClO3 Next, convert to moles of O2 by using the mole ratio between O2 and KClO3. Referring to the coefficients of the balanced equation, we get the mole ratio 2:3. Set it up in your equation like this: 122 g KClO3 * 1 mol KClO3 / 122.55 g KClO3 * 3 mol O2 / 2 mol KClO3 Convert now to grams of O2. Since you’re asked the amount of oxygen gas, you need to use 32 g/mol. If otherwise, just use 16 g/mol, the molar mass of O. 122 g KClO3 * 1 mol KClO3 / 122.55 g KClO3 * 3 mol O2 / 2 mol KClO3 * 32 g O2 / 1 mol O2 Multiply now the thing by the density.
