Question

Potassium chlorate, KClO3, decomposes when heated to produce potassium chloride and oxygen gas. If 2.50 grams of KCLO3 were heated in a test tube, how many grams of oxygen gas should be given off?

Answer 1

Answer:

0.98 g $O_{2}$

Explanation:

• $O_{2}$ = 2.5 g KCLO3 x $\frac{1 mol KCLO3}{122.5 g KCLO3} * \frac{3 mol O2}{2 mol KCLO3} * \frac{32 g O2}{1 mol O2} = 0.98 g O$2

Answer 2

Potassium chlorate has a molas mass of 122.55 g/mol. So, 2.50 g of KClO3 is,
2.5 g / (122.55 g/mol) = 0.0204 moles KClO3

The balanced chemical reaction is this:
2KClO3 ----> 2KCl + 3O2

So, for every 2 moles of KClO3, 3 moles of O2 forms. Using this ratio,
0.0204 moles KClO3 (3/2) = 0.0306 moles O2.

Converting moles of O2 to grams of O2:
0.0306 moles (16 g/mol) = 0.4896 grams

Therefore, around 0.49 grams of O2 will be formed if 2.50 g of KClO3 decomposed.