Calculate the volume of oxygen that would be produced if 12.35g of potassium trioxocarbonate iv is decmposed using:
2Kclo3➡2kcl + 3o2
Using stoichiometry
To calculate the volume of oxygen produced, we need to use stoichiometry to determine the moles of oxygen produced, and then convert it to volume using the ideal gas law.
1. Convert the mass of potassium trioxocarbonate (KClO3) to moles.
To do this, we need to know the molar mass of KClO3, which is:
Molar mass of K = 39.1 g/mol
Molar mass of Cl = 35.5 g/mol
Molar mass of O = 16.0 g/mol
So the molar mass of KClO3 = (39.1 + 35.5 + (3 * 16.0)) g/mol = 122.55 g/mol
Moles of KClO3 = 12.35 g / 122.55 g/mol = 0.1008 mol
2. Use the stoichiometry of the balanced chemical equation to determine the moles of oxygen produced.
From the balanced equation, we see that for every 2 moles of KClO3, 3 moles of O2 are produced.
Moles of O2 = (0.1008 mol KClO3) * (3 mol O2 / 2 mol KClO3) = 0.1512 mol
3. Convert moles of oxygen to volume.
In order to convert moles of oxygen to volume, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assume at STP, which is 1 atm)
V = volume (what we want to find)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (assume at STP, which is 273 K)
V = (nRT) / P
V = (0.1512 mol * 0.0821 L.atm/mol.K * 273 K) / 1 atm
Calculating this, we get:
V = 3.76 L
Therefore, the volume of oxygen produced is approximately 3.76 L.