Nitrous Oxide(N2O), or laghing gas, is mommonly used as a anesthetic in denistry and surgery. How many moles are present in 8.4 liters of nitrous oxide at STP?
0.375 is the answer
1 mole of any gas @ STP occupies 22.4L. Your sample is 8.4L. How many moles is that?
One mole of any gas at standard temperature and pressure (STP) is equal to 22.4 liters. To convert from liters to moles, simply multiply the given value (in liters) by the conversion factor (1 mol gas/22.4 L gas).
(8.4 L N2O × 1 mol N2O) / (22.4 L N2O) = 0.375 moles N2O
To determine the number of moles present in a given volume of a gas at Standard Temperature and Pressure (STP), you need to use the Ideal Gas Law. The Ideal Gas Law equation is as follows:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal Gas Constant
T = Temperature
At STP, the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K). The value of the ideal gas constant (R) is 0.0821 L·atm/(mol·K).
In this case, you are given that the volume of nitrous oxide (N2O) is 8.4 liters (L) at STP. Therefore, substituting the given values into the equation:
(1 atm) * (8.4 L) = n * (0.0821 L·atm/(mol·K)) * (273.15 K)
Solving this equation will give you the value of n (number of moles).
(1 atm) * (8.4 L) = n * (0.0821 L·atm/(mol·K)) * (273.15 K)
8.4 = n * (0.0821) * (273.15)
n = 8.4 / ((0.0821) * (273.15))
Calculating this expression gives us:
n ≈ 0.334 moles
Therefore, approximately 0.334 moles of nitrous oxide (N2O) are present in 8.4 liters at STP.