You are planning an experiment that requires 0.0610 mol of nitrogen monoxide gas (NO). What volume would you need at STP?
Answer in units of L
To find the volume of nitrogen monoxide gas (NO) required at STP (Standard Temperature and Pressure), you can use the ideal gas law equation:
PV = nRT
where:
P = pressure (STP = 1 atm)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP = 273.15 K)
Substituting the given values:
n = 0.0610 mol
P = 1 atm
R = 0.0821 L·atm/mol·K
T = 273.15 K
The equation becomes:
(1 atm) * V = (0.0610 mol) * (0.0821 L·atm/mol·K) * (273.15 K)
Simplifying:
V = (0.0610 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / 1 atm
V ≈ 1.567 L
Therefore, you would need approximately 1.567 L of nitrogen monoxide gas (NO) at STP.
To find the volume of nitrogen monoxide gas (NO) at STP (Standard Temperature and Pressure), we need to know the molar volume of an ideal gas at STP.
The molar volume at STP is defined as 22.4 liters per mole. This value is derived from the ideal gas law and is true for any ideal gas at STP.
Given that we have 0.0610 moles of NO gas, we can use the molar volume at STP to determine the volume of gas needed.
Volume (V) = Number of moles (n) × Molar volume at STP
V = 0.0610 mol × 22.4 L/mol
V ≈ 1.3664 L
Therefore, you would need approximately 1.3664 liters of nitrogen monoxide gas (NO) at STP for your experiment.
PV=nRT
solve for V, knowing P, n, R, T
PV=nRT
solve for V, knowing P, n, R, T
or, V=.0610*22.4 liters