Potassium chlorate decomposes according to the following equation:
2KCIO3(s)= 2KCl(s) + 3O2(g)
what volume of a gas can be produced by the decomposition of 12.6g of potassium chlorate measured under the following conditions?
a) at STP
b) at SATP
mols KClO3 = grams/molar mass = ?
For every 2 mols KClO3 you get 3 mol O2.
1 mol O2 @ STP occupies 22.4 L so ? mols will give ......
1 mol O2 @ SATP occup;ies 24.5 L so ? mols will give .....
Post your work if you get stuck..
To calculate the volume of gas produced by the decomposition of potassium chlorate, you need to use the ideal gas law equation: PV = nRT.
a) STP (Standard Temperature and Pressure):
At STP, the temperature is 273.15 K (0 degrees Celsius) and the pressure is 1 atmosphere (atm).
1. First, we need to convert the mass of potassium chlorate (KCIO3) to moles. The molar mass of KCIO3 is:
Potassium (K) = 39.10 g/mol
Chlorine (Cl) = 35.45 g/mol
Oxygen (O) = 16.00 g/mol
So, the molar mass of KCIO3 = 39.10 + 35.45 + (16.00 x 3) = 122.55 g/mol.
To find the number of moles (n), divide the mass of potassium chlorate (12.6 g) by its molar mass:
n = 12.6 g / 122.55 g/mol = 0.1028 mol
2. Since the balanced equation shows that 2 moles of KCIO3 produce 3 moles of O2, we can also determine the number of moles of O2 produced:
0.1028 mol KCIO3 x (3 mol O2 / 2 mol KCIO3) = 0.1542 mol O2
3. Now, we can use the ideal gas law equation to calculate the volume (V) of the gas:
PV = nRT
V = (nRT) / P
Substituting the values into the equation:
V = (0.1542 mol) x (0.0821 L × atm / mol × K) x (273.15 K) / (1 atm)
V = 0.1542 mol x 22.4 L/mol
V = 3.46 L
Therefore, at STP (0 degrees Celsius and 1 atm), the volume of gas produced by the decomposition of 12.6 g of potassium chlorate is 3.46 liters.
b) SATP (Standard Ambient Temperature and Pressure):
At SATP, the temperature is usually 298 K (25 degrees Celsius), and the pressure is 1 atmosphere (atm).
The calculation for SATP is similar to the one for STP, except we will use the SATP temperature of 298 K instead of 273.15 K.
Repeat steps 1 to 3 using the SATP temperature (298 K) instead of 273.15 K to find the volume of gas produced at SATP.
V = (0.1542 mol) x (0.0821 L × atm / mol × K) x (298 K) / (1 atm)
V = 0.1542 mol x 24.51 L/mol
V = 3.78 L
Therefore, at SATP (25 degrees Celsius and 1 atm), the volume of gas produced by the decomposition of 12.6 g of potassium chlorate is 3.78 liters.