When heated, KClO3 decomposes into KCl and O2. If this reaction produced 79.9 g of KCl, how much O2 was produced (in grams)?
2 KClO3 ---> 2 KCl + 3 O2
moles O2 = 3/2 * moles KCl
moles KCl = mass KCl / molar mass KCl
mass O2 = moles O2 * molar mass O2
To find out how much O2 was produced, we can start by finding the molar mass of KClO3 and KCl.
The molar mass of KClO3 is:
(1 atom of K) + (1 atom of Cl) + (3 atoms of O) = (39.1 g/mol) + (35.5 g/mol) + (3 * 16.0 g/mol) = 122.55 g/mol.
The molar mass of KCl is:
(1 atom of K) + (1 atom of Cl) = (39.1 g/mol) + (35.5 g/mol) = 74.6 g/mol.
Since KClO3 decomposes into KCl and O2, the molar mass of KClO3 is equal to the sum of the molar masses of KCl and O2:
122.55 g/mol = 74.6 g/mol + molar mass of O2.
To find the molar mass of O2, we subtract the molar mass of KCl from the molar mass of KClO3:
molar mass of O2 = 122.55 g/mol - 74.6 g/mol = 47.95 g/mol.
Now we need to find the number of moles of KCl produced, which is given by the mass of KCl produced divided by its molar mass:
moles of KCl = 79.9 g / 74.6 g/mol = 1.071 mol.
Since the reaction produces an equal number of moles of O2, we can find the mass of O2 by multiplying the number of moles of O2 by its molar mass:
mass of O2 = moles of O2 * molar mass of O2 = 1.071 mol * 47.95 g/mol = 51.37 g.
Therefore, approximately 51.37 grams of O2 were produced in the reaction.