Question

What volume of oxygen at stp is produced on heating 42.02 g of potassium trioxochlo rate v

Answer 1

To determine the volume of oxygen produced, we need to use the molar ratio between potassium trioxochlorate(V) and oxygen.

First, let's calculate the moles of potassium trioxochlorate(V) using its molar mass:

Molar mass of potassium trioxochlorate(V) (KClO₃) = 122.55 g/mol

Moles of KClO₃ = mass of KClO₃ / molar mass of KClO₃
= 42.02 g / 122.55 g/mol
= 0.342 moles

According to the balanced chemical equation for the decomposition of KClO₃:

2 KClO₃ → 2 KCl + 3 O₂

The molar ratio between KClO₃ and O₂ is 3:1. Therefore, for every 2 moles of KClO₃, 3 moles of O₂ are produced.

Moles of O₂ produced = (3 moles O₂ / 2 moles KClO₃) x 0.342 moles KClO₃
= 0.513 moles

Now, we can use Avogadro's law to determine the volume of oxygen at STP (standard temperature and pressure):

1 mole of any ideal gas at STP occupies 22.4 L.

Volume of O₂ produced = moles of O₂ produced x 22.4 L/mol
= 0.513 moles x 22.4 L/mol
= 11.48 L

Therefore, the volume of oxygen produced on heating 42.02 g of potassium trioxochlorate(V) at STP is approximately 11.48 liters.