Calculate the mass and volume of oxygen released when 29.4g pottasium chlorate decomposes to form pottasium chloride and oxygen
2KClO3 ==> 2KCl + 3O2
moles KClO3 = g/molar mass = 29.4/122.6 = approx 0.24
2 mols KClO3 will produce 3 mols O2; thereofre, 0.24 mols KClO3 will produce 0.24 x 3/2 = approx 0.36
1 mol O2 will occupy a volume of 22.4 L @ STP so 0.36 mols will have a volume of .......?
mass O2 = moles x molar mass = ?
2KClO3(s)>2KCl(s)+3O2(g)
RFM of KClO3 =39+35.5+3×16 =122.5
moles of KClO3 =29.4/122.5 =0.33moles
mole ratio of KClO3 to O2 is 2:3
moles of O2 =3/2×0.33= 0.495moles
mass=moles×molar mass
=0.495×16
=7.92g
1 mole at STP=22.4dm^3
0.495 moles=??
=0.495×22.4
=11.088dm^3
To calculate the mass and volume of oxygen released when potassium chlorate decomposes, we need to use the balanced chemical equation for the decomposition reaction of potassium chlorate:
2KClO3 -> 2KCl + 3O2
From the equation, we can see that 2 moles of potassium chlorate (KClO3) produce 3 moles of oxygen gas (O2).
First, we convert the given mass of potassium chlorate (29.4g) to moles using its molar mass. The molar mass of KClO3 is calculated by adding the atomic masses of each element: K (39.10 g/mol) + Cl (35.45 g/mol) + 3O (16.00 g/mol) = 122.55 g/mol.
moles of KClO3 = mass / molar mass = 29.4g / 122.55 g/mol ≈ 0.24 mol
Next, we use the stoichiometry of the balanced equation to determine the moles of oxygen gas released.
moles of O2 = 3 moles of O2 / 2 moles of KClO3 × 0.24 mol = 0.36 mol
Finally, we can calculate the mass and volume of oxygen using the molar mass of oxygen (O2) which is 32 g/mol, and the ideal gas law where the molar volume of a gas is 22.4 L/mol at standard temperature and pressure (STP).
Mass of O2 = moles of O2 × molar mass = 0.36 mol × 32 g/mol = 11.52 g
Volume of O2 = moles of O2 × molar volume at STP = 0.36 mol × 22.4 L/mol = 8.06 L
Therefore, the mass of oxygen released is approximately 11.52 grams, and the volume of oxygen released is approximately 8.06 liters.
To calculate the mass and volume of oxygen released when potassium chlorate decomposes, we need to use the molar ratios of the reactants and products involved in the reaction.
First, let's write the balanced chemical equation for the decomposition of potassium chlorate:
2KClO3(s) -> 2KCl(s) + 3O2(g)
According to this equation, 2 moles of potassium chlorate produce 3 moles of oxygen gas.
Step 1: Calculate the number of moles of potassium chlorate (KClO3) present in 29.4g:
To do this, we need to know the molar mass of KClO3. Potassium (K) has a molar mass of 39.10 g/mol, Chlorine (Cl) has a molar mass of 35.45 g/mol, and Oxygen (O) has a molar mass of 16.00g/mol.
Molar mass of KClO3 = (1 * 39.10) + (1 * 35.45) + (3 * 16.00) = 122.55 g/mol
Number of moles of KClO3 = mass / molar mass
= 29.4 g / 122.55 g/mol
≈ 0.240 mol
Step 2: Use the molar ratio to determine the number of moles of oxygen (O2):
We know from the balanced equation that the molar ratio between KClO3 and O2 is 2:3.
Moles of O2 = 0.240 mol KClO3 * (3 mol O2 / 2 mol KClO3)
= 0.360 mol O2
Step 3: Calculate the mass of oxygen (O2):
To find the mass, we need to multiply the number of moles by the molar mass of oxygen.
Mass of O2 = moles * molar mass
= 0.360 mol * 32.00 g/mol (molar mass of O2)
= 11.52 g
So, the mass of oxygen released during the decomposition of 29.4g of potassium chlorate is 11.52g.
Step 4: Calculate the volume of oxygen (O2) at standard temperature and pressure (STP):
At STP, 1 mole of any gas occupies 22.4 liters of volume.
Volume of O2 = moles * 22.4 L/mol
= 0.360 mol * 22.4 L/mol
= 8.06 L (rounded to two decimal places)
Therefore, the volume of oxygen released during the decomposition of 29.4g of potassium chlorate is 8.06 liters at STP.