calculate the volume of oxygen at 15 degree Celsius and 745mmHg pressure. which could be obtained by heating 10g of potassium trioxochlorate(v)
2KClO3 ==> 2KCl + 3O2
mols KClO3 = 10/molar mass KClO3 = ?
Use the coefficients in the balanced equation to convert mols KClO3 to mols O2. That's mols O2 at STP = ?mols KClO3 x 3 mols O2/2 mol KClO3 = ? mols KClO3 x 3/2 = ?
The use PV = nRT to convert mols O2 at STP to mols O2 at the conditions of the problem.
To calculate the volume of oxygen, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L*atm/mol*K)
T = temperature in Kelvin
First, we need to convert the given temperature to Kelvin:
15 degrees Celsius + 273.15 = 288.15 K
Next, we convert the given pressure from mmHg to atm:
745 mmHg ÷ 760 mmHg/atm = 0.979 atm
Now, we need to calculate the number of moles of oxygen produced by heating 10g of potassium trioxochlorate(V). The balanced chemical equation for the decomposition of potassium trioxochlorate(V) is:
2KClO3 → 2KCl + 3O2
The molar mass of potassium trioxochlorate(V) (KClO3) is:
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16.0 g/mol
Molar mass of KClO3 = 39.1 g/mol + (35.5 g/mol + 16.0 g/mol + 16.0 g/mol) = 122.6 g/mol
Using the molar mass, we can calculate the number of moles:
moles of KClO3 = mass (g) / molar mass (g/mol)
moles of KClO3 = 10 g / 122.6 g/mol ≈ 0.0816 mol
According to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2. Therefore, the number of moles of oxygen produced is:
moles of O2 = (0.0816 mol KClO3) × (3 mol O2 / 2 mol KClO3) ≈ 0.1224 mol O2
Now, we can plug the values into the ideal gas law equation to calculate the volume:
PV = nRT
(0.979 atm)(V) = (0.1224 mol)(0.0821 L*atm/mol*K)(288.15 K)
V = (0.1224 mol × 0.0821 L*atm/mol*K × 288.15 K) / 0.979 atm
V ≈ 2.23 L
Therefore, the volume of oxygen produced is approximately 2.23 liters.