Question

How many grams of oxygen are produced when 6.06 g of potassium chlorate decompose completely

Answer 1

2.38 g of oxygen (O2).

Step-by-step explanation:

What is given?

Mass of potassium chlorate (KClO3) = 6.06 g.

Molar mass of KClO3 = 122.4 g/mol.

Molar mass of oxygen (O2) = 32 g/mol.

Step-by-step solution:

First, let's state the balanced chemical equation. Remember that the decomposition of a compound produces two or more products:

`2KClO_3\rightarrow2KCl+3O_2.`

Now, let's convert 6.06 g of KClO3 to moles using its molar mass:

`6.06\text{ g KClO}_3\cdot\frac{1\text{ mol KClO}_3}{122.4\text{ g KClO}_3}=0.0495\text{ moles KClO}_3.`

You can see in the chemical equation that 2 moles of KClO3 produce 3 moles of O2. By doing a rule of three with this data, we obtain that:

`0.0495\text{ moles KClO}_3\cdot\frac{3\text{ moles O}_2}{2\text{ moles KClO}_3}=0.0743\text{ moles O}_2.`

The final step is to convert from 0.0743 moles of O2 to grams using its molar mass, like this:

`0.0743\text{ moles O}_2\cdot\frac{32\text{ g O}_2}{1\text{ }mol\text{ }O_2}=2.38\text{ g O}_2.`

The answer is that we will produce 2.38 g of oxygen (O2) from the decomposition of 6.06 g of potassium chlorate (KClO3).