Respuesta :
Answer:
0.217 mol
Explanation:
Given:
Temperature = 14° C + 273 = 287 K.
Atmospheric pressure = 790 torr
Volume = 5 L
Vapor pressure of water = 12 torr
Now,
The vapor pressure of the Oxygen from the Law of Partial Pressures
= 790 torr - 12 torr
= 778 torr
Converting in terms of atm
we know,
1 atm = 760 torr
therefore,
778 torr = [tex]\frac{1}{760}\times778[/tex] = 1.02368 atm.
also,
PV = nRT
where,
P is the pressure
V is the volume
n is number of moles
R is the gas constant = 0.0821 L-atm/mol-K
T is the temperature
on substituting the respective values, we get
1.02368 × 5 = n × 0.0821 × 287
or
n = 0.217 mol
Based on the ideal gas equation, the amount of oxygen gas collected is 0.217 moles
What is the amount of gas collected?
The amount of a gas present in a given volume of gas at a given temperature and pressure is determined using the ideal gas equation.
The ideal gas equation is given as:
- PV = nRT
where;
- n is the amount of gas in moles
- R is molar gas constant = 0.0821 L-atm/mol-K
- T is temperature in Kelvin
- P is pressure
- V is volume
Temperature = 14° C + 273 = 287 K.
Atmospheric pressure = 790 torr
Volume = 5 L
Vapor pressure of water = 12 torr
Vapor pressure of the Oxygen = 790 torr - 12 torr
Vapor pressure of the Oxygen = 778 torr
Converting torr to atm
1 atm = 760 torr
778 torr = = 1.02368 atm.
From Ideal gas equation
- n = PV/RT
n = 1.02368 × 5/0.0821 × 287
n = 0.217 moles
Therefore, the amount of oxygen gas collected is 0.217 moles
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