What volume of O2 can be made from 5.00 x 10-2 moles of KClO3?
The balanced chemical equation for the decomposition of potassium chlorate (KClO3) is:
2 KClO3 -> 2 KCl + 3 O2
According to the equation, every 2 moles of KClO3 produce 3 moles of O2.
So, if we have 5.00 x 10^-2 moles of KClO3, we can calculate the volume of O2 produced.
Number of moles of O2 = (5.00 x 10^-2 moles KClO3) x (3 moles O2 / 2 moles KClO3)
= 7.50 x 10^-2 moles of O2
Assuming ideal gas behavior, we can use the ideal gas law to calculate the volume of O2:
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)
We'll assume standard temperature and pressure (STP), which is 1 atmosphere (atm) and 273 Kelvin (K).
Using the ideal gas law, we can rearrange the formula to solve for volume:
V = (nRT) / P
V = (7.50 x 10^-2 moles) x (0.0821 L.atm/mol.K) x (273 K) / 1 atm
V ≈ 1.95 liters
Therefore, approximately 1.95 liters of O2 can be made from 5.00 x 10^-2 moles of KClO3.