when heated, kclo3 decomposes into kcl ad 02. 2kclo3--> 2kcl+3o2
if this reaction produced 61.7 g of kcl, how much o2 was produced (in grams)?
one gets 1.5 moles O2 for each mole of KCl.
moles KCl=61.7/formulamassKCL
moles O2= 1.5 times that
mass O2=moles O2 * 32
To determine the amount of O2 produced when 61.7 g of KCl is formed in the reaction, we need to use the stoichiometry of the balanced chemical equation.
According to the balanced equation: 2KClO3 → 2KCl + 3O2
The molar ratio between KClO3 and O2 is 2:3. This means that for every 2 moles of KClO3, 3 moles of O2 are produced.
Step 1: Calculate the number of moles of KCl formed:
Molar mass of KCl = 39.10 g/mol (potassium) + 35.45 g/mol (chlorine) = 74.55 g/mol
Number of moles of KCl = Mass of KCl / Molar mass of KCl
Number of moles of KCl = 61.7 g / 74.55 g/mol = 0.8272 mol (rounded to four decimal places)
Step 2: Use the stoichiometry of the balanced equation to find the number of moles of O2 produced:
According to the balanced equation, the ratio of KClO3 to O2 is 2:3.
Number of moles of O2 = (Number of moles of KClO3) x (3 moles of O2 / 2 moles of KClO3)
Number of moles of O2 = 0.8272 mol x (3 mol / 2 mol) = 1.2408 mol (rounded to four decimal places)
Step 3: Calculate the mass of O2 produced:
Molar mass of O2 = 16.00 g/mol (oxygen) x 2 = 32.00 g/mol
Mass of O2 = Number of moles of O2 x Molar mass of O2
Mass of O2 = 1.2408 mol x 32.00 g/mol = 39.70 g (rounded to two decimal places)
Therefore, approximately 39.70 grams of O2 was produced in the reaction.