What mass of KClO3 must be decomposed to KCl and O2 to give 638ml O2 at 18 degrees Celcius and 752torr?
Here is a worked example using the same material.
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To determine the mass of KClO3 that must be decomposed, we need to use the ideal gas law and stoichiometry.
First, let's establish the given conditions:
- Volume of O2 gas (V) = 638 mL = 0.638 L
- Temperature (T) = 18 degrees Celsius = 18 + 273.15 Kelvin = 291.15 K
- Pressure (P) = 752 torr
Next, let's convert the pressure unit from torr to atmospheres (atm):
1 atm = 760 torr
Therefore, P = 752 torr * (1 atm / 760 torr) ≈ 0.9895 atm
Now, we can 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
Rearranging the equation to solve for the number of moles (n):
n = PV / RT
Substituting the given values:
n = (0.9895 atm) * (0.638 L) / (0.0821 L·atm/mol·K) * (291.15 K)
Calculating n:
n ≈ 0.01646 moles of O2 gas
Now, we can use the balanced chemical equation for the decomposition of KClO3:
2 KClO3 → 2 KCl + 3 O2
From the equation, we can see that for every 3 moles of O2 gas produced, 2 moles of KClO3 are decomposed. Therefore, the molar ratio of O2 to KClO3 is 3:2.
Using this ratio, we can calculate the number of moles of KClO3:
0.01646 moles O2 * (2 moles KClO3 / 3 moles O2) ≈ 0.01097 moles KClO3
Finally, to find the mass of KClO3, we need to use its molar mass.
The molar mass of KClO3 is:
K (39.10 g/mol) + Cl (35.45 g/mol) + 3 O (16.00 g/mol) = 122.55 g/mol
Calculating the mass of KClO3:
0.01097 moles KClO3 * (122.55 g/mol) ≈ 1.347 g
Therefore, approximately 1.347 grams of KClO3 must be decomposed to produce 638 mL of O2 gas at 18 degrees Celsius and 752 torr.