By using stoichiometry, 0.320 grams of oxygen gas corresponds to 4.01 x 10²° molecules of KCl being produced, based on the balanced chemical equation of potassium chlorate decomposition.
To determine how many molecules of KCl are produced when 0.320 grams of oxygen are also created, we need to use stoichiometry. Given that 1 mol of O₂ has a mass of 32.00 g/mol, we can calculate the moles of O₂:
0.320 g O₂ × (1 mol O₂ / 32.00 g O₂) = 0.010 mol O₂
Potassium chlorate (KClO₃) decomposes to produce potassium chloride (KCl) and oxygen gas (O₂). The balanced equation for this reaction is:
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
This tells us that 2 moles of KClO₃ produce 3 moles of O₂. Using the stoichiometry of the reaction, we find that:
0.010 mol O₂ × (2 mol KCl / 3 mol O₂) = 0.00667 mol KCl
Finally, we convert moles of KCl to molecules:
0.00667 mol KCl × (6.022 × 10²³ molecules/mol) = 4.01 × 10²° molecules of KCl