How many grams of oxygen (O2) are formed when 58.6 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11 g/mol. 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)

Question

Formula: 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)

Answer 1

Your given is 58.6 g KNO_3, so you will start there and end in oxygen. This is a grams to grams problem, so 7 boxes total will be used or 3 full conversion factors.

The set up would be as follows:

58.6 g KNO₃ x (1 mol KNO₃ /101.11 g KNO₃) x ( 5 mol O₂ / 4 mol KNO₃ ) x (31.998 g O₂ / 1 mol O₂)

From your given, you'd use the molar mass to convert from g KNO₃ to mol KNO_3. The next step requires a mole ratio to convert from mol KNO₃ to mol O_2. Finally, to get from mol of O₂ to grams of O₂, you'd multiply by the molar mass of O₂

58.6 g x (1/101.11 g) x (1.25) x (31.998) =

Answer = 23.18 g O₂