What mass of potassium chloride would be formed when 7.5moles of potassium carbonate iv is used.
2Kclo3➡2kcl + 3o2
Using stoichiometry
To determine the mass of potassium chloride formed, we need to use the stoichiometry of the reaction.
From the balanced equation:
2KClO3 → 2KCl + 3O2
We can see that for every 2 moles of potassium chlorate (KClO3), we get 2 moles of potassium chloride (KCl).
Given that we have 7.5 moles of potassium carbonate (K2CO3), we need to convert this to moles of potassium chloride (KCl) using stoichiometry.
Since the molar ratio between K2CO3 and KClO3 is 1:1, we have:
7.5 moles K2CO3 × (2 moles KClO3 / 1 mole K2CO3) = 15 moles KClO3
Since the molar ratio between KClO3 and KCl is 2:2, we have:
15 moles KClO3 × (2 moles KCl / 2 moles KClO3) = 15 moles KCl
Now, we can calculate the mass of potassium chloride using the molar mass of KCl (39.09 g/mol):
15 moles KCl × (39.09 g KCl / 1 mole KCl) = 586.35 g KCl
Therefore, 586.35 grams of potassium chloride would be formed when 7.5 moles of potassium carbonate are used.