How many mole of KClO3 are needed to make 320g O2
2KClO3 ==> 2KCl + 3O2
How many mols O2 in 320 g? That's mol = grams/molar mass = ?
Then 2 mols KClO3 = 3 mols O2; therefore,
.....x mols KClO3 = ? mols O2.
To determine the number of moles of KClO3 needed to produce 320g of O2, you would need to use stoichiometry and the balanced chemical equation of the reaction.
The balanced chemical equation for the decomposition of KClO3 is:
2KClO3 -> 2KCl + 3O2
According to the equation, for every 2 moles of KClO3, 3 moles of O2 are produced.
First, calculate the molar mass of O2:
2(atomic mass of O) = 2(16.00 g/mol) = 32.00 g/mol
Next, calculate the number of moles of O2 using the given mass:
moles of O2 = mass of O2 / molar mass of O2
= 320 g / 32.00 g/mol
= 10 moles
Since the ratio of moles of KClO3 to moles of O2 is 2:3, you can set up a proportion to find the moles of KClO3:
2 moles KClO3 / 3 moles O2 = x moles KClO3 / 10 moles O2
Cross-multiplying:
2 moles KClO3 * 10 moles O2 = 3 moles O2 * x moles KClO3
20 moles KClO3 = 3x moles KClO3
Dividing both sides by 3:
20 moles KClO3 / 3 = x moles KClO3
x ≈ 6.67 moles KClO3
Therefore, approximately 6.67 moles of KClO3 are needed to produce 320g of O2.