How many mols of dinitrogen monoxide are contained in 56.0L at STP?
56.0L x (1 mol/22.4L) = ? mols.
It is 2.50 moles if you are using significant digits!
56.0 L • 1 mol /22.4 L = 2.50 mol
To calculate the number of moles of a gas, you can use the ideal gas law equation: PV = nRT. However, in this case, we are given the volume (V), which is 56.0L, and the conditions are STP (Standard Temperature and Pressure).
At STP, the temperature (T) is 273.15 Kelvin (0 degrees Celsius) and the pressure (P) is 1 atmosphere.
First, we need to convert the volume to liters (L) to match the units of the ideal gas law equation. No conversion is needed in this case since the given volume is already given in liters.
Now we can substitute the known values into the equation:
PV = nRT
(1 atm) * (56.0 L) = n * (0.0821 L * atm/mol * K) * (273.15 K)
56.0 = n * 22.4
Now, we can solve for n:
n = (56.0 L) / (22.4 L/mol)
n = 2.500 mol
Therefore, there are 2.500 moles of dinitrogen monoxide contained in 56.0L at STP.