Question

What volume of O2 measured at STP will be produced by the decomposition of 6.75 g KClO3? Use the equation 2KClO3(s) → 2KCl(s) + 3O2(g).

Answer 1

Approximately 1.85 liters of O2 at STP.

Step-by-step explanation:

To solve this problem, we first need to use the molar mass of KClO3 to convert grams to moles. The molar mass of KClO3 is approximately 122.55 g/mol.

So, the number of moles of KClO3 is:

122.55g/mol6.75g​=0.0551mol

From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of O2. So, the number of moles of O2 produced is:

0.0551mol×23​=0.08265mol

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. So, the volume of O2 produced is:

0.08265mol×22.4L/mol=1.85L

So, the decomposition of 6.75 g of KClO3 will produce approximately 1.85 liters of O2 at STP.

Hope it Helps!