Use the following reaction: 2KClO3 -> 2KCl + 3O2

Question

If 5.00 g of KClO3 is decomposed, how many grams of O2 is produced?

Answer 1

This looks like a classic stoichiometry problem: a certain amount of a reactant is added, and it should theoretically generate a certain amount of product.

How do we know how much product? It'll depend on a couple things: molar ratios and molar mass.

Molar ratios describe the relationships between amounts of reactants and products in a balanced chemical reaction (mol of X/mol of Y). In this reaction, 2 moles of KClO₃ generates 3 moles of O₂ (and 4 moles would generate 6 moles, 4 molecules would generate 6 molecules, etc.).

Molar masses describe the relationship between a mole of a substance and its mass (g/mol). We get this information from the periodic table: the molar mass of KClO₃ is the molar mass of (K + Cl + 3(O)) in g/mol = 122.55 g/mol and the molar mass of O₂ is (2(O)) in g/mol = 32 g/mol.

We convert g of KClO₃ to g of O₂ by paying close attention to units:

1. 5.00 g KClO₃ needs to be converted to mol KClO₃ using molar mass so that we can use molar ratios in the next step: 5.00 g KClO₃ * (1 mol KClO₃ / 122.55 g KClO₃) = 0.0407997 mol KClO₃. Notice how grams of KClO₃ cancel and we end up in moles of KClO₃.
2. Next, 0.0407997 mol KClO₃ needs to convert into mol of O₂, the product of interest, using molar ratios from the balanced chemical reaction: 0.0407997 mol KClO₃ * (3 mol O₂/ 2 mol KClO₃) = 0.0611995 mol O₂. Here, we've canceled mol of KClO₃ and resulted in mol of O₂, the product of interest.
3. Lastly, we care about grams of O₂, not moles of O₂. We'll convert by multiplying the moles of O₂ by the molar mass of O₂: 0.0611995 mol O₂ * (32 g O₂/1 mol O₂) = 1.95838 g O₂. Again, notice that we cancel mol of O₂ and end up with g of O_2, the desired units.
4. We should change the number of significant figures to preserve the level of accuracy from the given mass. 5.00 g has 3 sig figs, and we use multiplication and division throughout our calculations, so our result should have 3 sig figs, too: 1.96 g O₂.

It's possible to combine all these steps using the "train track method," shown below with canceled units:

5.00 g KClO₃ * (1 mol KClO₃ / 122.55 g KClO₃) * (3 mol O₂/ 2 mol KClO₃) * (32 g O₂/1 mol O₂) = 1.96 g O₂