Calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate?

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

75.0 g KClO3 (1 mol KClO3 /122.55 g KClO3) (3 mol O2/2 mol KClO3) ( 32 g O2 / 1mol O2) = 58.75 g O2

Thus, the answer is 28.88g H2O.

29.4 g

Explanation:

Let's consider the decomposition of potassium chlorate.

KClO₃ → KCl + 1.5 O₂

The molar mass of KClO₃ is 122.55 g/mol. The moles of KClO₃ in 75.0 g are:

75.0 g × (1 mol/ 122.55 g) = 0.612 mol

The molar ratio of KClO₃ to O₂ is 1:1.5. The moles of O₂ are 1.5 × 0.612 mol = 0.918 mol

The molar mass of O₂ is 32.00 g/mol. The mass of O₂ corresponding to 0.918 moles is:

0.918 mol × (32.00 g/mol) = 29.4 g

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