when magnesium chlorate under goes decompostion, oxgen gas is realsed in the following reaction:
2KClO3→ 2KCl + 3O2
when 3.50 grams of potassuim chlorate is decomposed at STP, how many litters of oxgen will be produced?
at STP, a mol of perfect gas occupies 22.4 liters
so we need to know how many moles
How many grams in 1 mol of K Cl O3 ?
K = 39
Cl = 35.5
O3 = 16*3 = 48
---------------
mol = 122.6 grams
so
3.5 grams of KClO3 = .0286 mol
for every mol of KClO3 I get 3/2 mol of O2
so
(3/2) (.0286) = .0428 mol of O2
every mol is 22.4 liters of gas so
.0428 * 22.4 = .96 or about one liter of O2
I said 2KClO3
That is why I used (3/2)
To determine the number of liters of oxygen gas produced when 3.50 grams of potassium chlorate (KClO3) decomposes at STP (Standard Temperature and Pressure) using the given balanced equation:
2KClO3 → 2KCl + 3O2
We need to follow the following steps:
Step 1: Calculate the moles of potassium chlorate (KClO3)
The molar mass of KClO3 can be found in the periodic table:
Atomic mass of K = 39.10 g/mol
Atomic mass of Cl = 35.45 g/mol
Atomic mass of O = 16.00 g/mol
Total molar mass of KClO3 = (2 * atomic mass of K) + atomic mass of Cl + (3 * atomic mass of O)
Total molar mass of KClO3 = (2 * 39.10 g/mol) + 35.45 g/mol + (3 * 16.00 g/mol)
Total molar mass of KClO3 = 122.55 g/mol
Moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
Moles of KClO3 = 3.50 g / 122.55 g/mol = 0.0285 mol
Step 2: Use the stoichiometry of the balanced equation to convert moles of KClO3 to moles of oxygen (O2)
From the balanced equation:
2 moles of KClO3 produce 3 moles of O2
Moles of O2 = Moles of KClO3 * (3 moles of O2 / 2 moles of KClO3)
Moles of O2 = 0.0285 mol * (3/2) = 0.0428 mol
Step 3: Convert moles of oxygen (O2) to liters using the ideal gas law at STP (standard pressure and temperature)
At STP, 1 mole of any ideal gas occupies 22.4 liters.
Volume of O2 = Moles of O2 * 22.4 liters/mol
Volume of O2 = 0.0428 mol * 22.4 liters/mol = 0.95872 liters
Therefore, when 3.50 grams of potassium chlorate decomposes at STP, approximately 0.959 liters of oxygen gas will be produced.