Oxygen can be produced in the laboratory by the reaction
2 KClO3(s) → 2 KCl(s) + 3 O2(g) .
How much potassium chlorate is needed to produce 2.75 L of oxygen, collected over water at 37◦C and 94.9 kPa? The vapor pressure of water at 37◦C is 6.28 kPa.
1. 0.189 mol
2. 0.142 mol
3. 6.30 × 10−2 mol 4. 7.20 × 10−2 mol 5. 9.45 × 10−2 mol
Here is an example problem using KClO3 to product oxygen. Just follow the steps in reverse to find mols KClO3. Use PV = nRT to find mols O2 needed.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Why did the oxygen go to therapy? Because it heard potassium chlorate had some serious reaction skills!
To solve this problem, we need to use the ideal gas law to find the number of moles of oxygen. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
37°C + 273.15 = 310.15 K
Now we can use the ideal gas law to solve for the number of moles of oxygen:
PV = nRT
(94.9 kPa - 6.28 kPa) * (2.75 L) = n * (8.314 L·kPa/mol·K) * (310.15 K)
88.62 kPa * L = n * 2570.4 kPa·L·mol⁻¹·K⁻¹
n = (88.62 kPa * L) / (2570.4 kPa·L·mol⁻¹·K⁻¹)
n ≈ 0.03444 mol
But wait, we need to find the amount of potassium chlorate, not oxygen. According to the balanced equation, the stoichiometry is 2 moles of KClO3 produce 3 moles of O2. So, we need to multiply the number of moles of oxygen by the mole ratio:
n KClO3 = (0.03444 mol O2) * (2 mol KClO3 / 3 mol O2)
n KClO3 ≈ 0.02296 mol
So, the amount of potassium chlorate needed to produce 2.75 L of oxygen is approximately 0.02296 mol. But we need to choose the option closest to this value, so the correct answer is option 3. 6.30 × 10⁻² mol.
To find out how much potassium chlorate is needed to produce 2.75 L of oxygen, we can use the ideal gas law equation:
PV = nRT
Where:
P = total pressure (in this case, it's the pressure of the oxygen gas)
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature (in Kelvin)
Since the oxygen gas is collected over water, we need to take the vapor pressure of water into account. So, the total pressure (P) is equal to the pressure of oxygen gas plus the vapor pressure of water:
P = Poxygen + Pwater vapor
Using the given values:
Poxygen = 94.9 kPa
Pwater vapor = 6.28 kPa
P = 94.9 kPa + 6.28 kPa
P = 101.18 kPa
Converting the temperature from Celsius to Kelvin:
37°C + 273.15 = 310.15 K
Now we can rearrange the ideal gas law equation and solve for the number of moles (n) of oxygen gas:
n = PV / RT
n = (101.18 kPa) * (2.75 L) / [(0.0821 L·atm/(K·mol)) * (310.15 K)]
Simplifying:
n ≈ 0.142 mol
Therefore, the amount of potassium chlorate needed to produce 2.75 L of oxygen is approximately 0.142 mol. Hence, the correct answer is option 2. 0.142 mol.
To determine how much potassium chlorate is needed to produce 2.75 L of oxygen, collected over water at 37◦C and 94.9 kPa, we can use the ideal gas law.
The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
In this case, we need to take into account the pressure of water vapor. The total pressure in the system is the sum of the pressures of the oxygen gas and the water vapor:
P_total = P_oxygen + P_water
Given:
V = 2.75 L (volume of oxygen)
P_total = 94.9 kPa (total pressure)
P_water = 6.28 kPa (pressure of water vapor)
T = 37°C (temperature)
First, we need to convert the temperature from Celsius to Kelvin:
T = 37 + 273.15 = 310.15 K
Now, let's calculate the pressure of oxygen gas alone:
P_oxygen = P_total - P_water
P_oxygen = 94.9 kPa - 6.28 kPa
P_oxygen = 88.62 kPa
Next, using the ideal gas law, we can calculate the number of moles of oxygen gas:
PV = nRT
n = PV / RT
n = (88.62 kPa x 2.75 L) / (8.31 L·kPa/mol·K x 310.15 K)
n = 0.109 mol
However, the balanced chemical equation shows that 2 moles of potassium chlorate produce 3 moles of oxygen. So, we need to adjust the number of moles of oxygen to account for the stoichiometry of the reaction.
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Therefore, the number of moles of potassium chlorate needed is:
moles of KClO3 = (moles of O2 x 2) / 3
moles of KClO3 = (0.109 mol x 2) / 3
moles of KClO3 = 0.072 mol
So, the answer is option 4. 0.072 mol.