At STP, what is the volume of 8.38 mol of
nitrogen gas?
Answer in units of L
8.38 mol N2 x (22.4L/mol) = ?
To calculate the volume of a gas at STP (Standard Temperature and Pressure), you can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (which is 1 atm at STP)
V is the volume
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (which is 273.15 K at STP)
Rearranging the equation to solve for V, we get:
V = (nRT) / P
Now, we can substitute the given values into the equation:
n = 8.38 mol
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
P = 1 atm
V = (8.38 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
By simplifying the equation, we find:
V = 186.45 L
Therefore, the volume of 8.38 mol of nitrogen gas at STP is 186.45 L.