How much oxygen will we get from 120 grams of KClO3?
none, unless you decompose it.
With this equation
2KClO3>>>2KCl + 3O2
for each mole of KClO3, you get 1.5 moles O2.
you started with molesKClO3 as 120/molmassKClO3
multiply that by 1.5 to get moles O2
multipy then by 32 to get grams of O2
To determine how much oxygen can be obtained from 120 grams of KClO3 (potassium chlorate), we need to consider the stoichiometry of the chemical reaction.
First, let's write the balanced chemical equation for the decomposition of KClO3 into oxygen gas (O2) and potassium chloride (KCl):
2KClO3 -> 2KCl + 3O2
From the balanced equation, we can see that for every 2 moles of KClO3, we will obtain 3 moles of O2.
To calculate the number of moles of KClO3, we need to use the molar mass of KClO3, which is approximately 122.55 g/mol.
Moles of KClO3 = mass of KClO3 / molar mass of KClO3
= 120 g / 122.55 g/mol
≈ 0.979 moles
From the stoichiometry of the balanced equation, we know that 2 moles of KClO3 yield 3 moles of O2, so we can set up a proportion to calculate the moles of O2:
(3 moles O2 / 2 moles KClO3) = (x moles O2 / 0.979 moles KClO3)
Solving for x, we get:
x = (3 moles O2 / 2 moles KClO3) * 0.979 moles KClO3
= 1.4685 moles O2
Finally, to determine the mass of oxygen, we multiply the number of moles by the molar mass of oxygen, which is approximately 32 g/mol.
Mass of O2 = moles of O2 * molar mass of O2
= 1.4685 moles * 32 g/mol
≈ 46.956 g
Therefore, from 120 grams of KClO3, we can obtain approximately 46.956 grams of oxygen gas.