Question

If 5.0 g of potassium chlorate (KClO3) is decomposed, what volume of oxygen gas is produced at STP?

Answer 1

1.37dm³

Step-by-step explanation:

To solve this problem, let us write the reaction expression:

2KClO₃ → 2KCl + 3O₂

Now, mass of KClO₃ is 5g, let us find the number of moles;

Number of moles =
`(mass)/(molar mass)`

Molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

Now;

Number of moles =
`(5)/(122.5)` = 0.04mole

So;

2 moles of KClO₃ will produce 3 moles of oxygen gas

0.04 mole of KClO₃ will produce
`\frac{3 x 0.04} {2}` = 0.06moles

At STP;

1 mole of gas has a volume of 22.4dm³

0.06 mole of oxygen gas will have a volume of 22.4 x 0.06 = 1.37dm³