Chemistry Lab Question

Question

Assume that you carried out today’s reaction, but didn’t measure the masses of either the KClO3 starting material or the KCl product. However, you were able to determine that 0.65L of O2 gas was released. If 1 mol of O2 occupies 22.4L, what were the original masses of KClO3 and KCl used and produced in the reaction?

Answer 1

Anna,

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The reaction that you probably carried out was the decomposition of KClO3 as follows:

2KClO₃ ==> 2KCl + 3O₂ ... balanced equation

Since you determined that you had 0.65 L of O₂ (the product) and we know that 1 mol O2 occupies 22.4 L, we can now find the moles of O₂ produced.

0.65 L x 1 mol O2/22.4 L = 0.0290 moles of O₂

From the balanced equation, we see that 2 moles of KClO₃ produces 3 moles of O₂ and 2 moles of KCl. From this information, and using dimensional analysis, we can find mass of KClO₃ used and mass of KCl produced.

Molar mass KClO₃ = 123 g/mole

Molar mass KCl = 74.5 g/mole

mass of KClO₃ used: 0.0290 moles O₂ x 2 moles KClO₃ / 3 moles O₂ x 123 g/mol = 2.4 g KClO₃ (2 sig.figs.)

mass KCl produced: 0.0290 moles O₂ x 2 mol KCl / 3 mol O₂ x 74.5 g/mol = 1.4 g KCl (2 sig. figs.)