How many moles of O2 are present in 67.2 L of O2 gas at STP?
using equation PV=nRT. and the STP of T=273K and P=1atm. You have your V=67.2L and R is always .0821(atm/L)(mol/K). Your equation will end up to be n=PV/RT. And if I calculated rite your mole will be rounded to 3. But please check
As we know the no of moles at STP =volume in lt at STP/22.4lt so as per this,
67.2/22.4=3 moles
To determine the number of moles of O2 gas present at STP in a given volume, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
At STP (Standard Temperature and Pressure), the pressure is 1 atm, and the temperature is 273.15 K.
Given:
Volume (V) = 67.2 L (liters)
Pressure (P) = 1 atm
Temperature (T) = 273.15 K
Let's substitute the values into the ideal gas law equation:
PV = nRT
(1 atm)(67.2 L) = n(0.0821 L·atm/(mol·K))(273.15 K)
Now we can solve for the number of moles (n):
n = (1 atm)(67.2 L) / (0.0821 L·atm/(mol·K))(273.15 K)
Simplifying the expression:
n = (67.2) / (0.0821)(273.15)
n = 2.52 moles
Therefore, there are approximately 2.52 moles of O2 gas present in 67.2 L of O2 gas at STP.