Calculate the Volume of Ammonia Produced
How can we calculate the volume of ammonia produced?
Given data:
Mass of Mg₃N₂ = 6.0 g
Molar mass of Mg₃N₂ = 148.3 g/mol
STP conditions
Pressure (P) = 1 atm
Temperature (T) = 273.15 K
Ideal Gas Constant (R) = 0.082 L·atm/mol·K
From the reaction: Mg₃N₂ + 6H₂O -> 3Mg(OH)₂ + 2NH₃
We have calculated the moles of ammonia (NH₃) produced to be 0.081 mol. Now, how do we determine the volume of ammonia produced?
Answer:
To calculate the volume of ammonia produced, we can use the ideal gas equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal Gas Constant
T = Temperature
Calculation:
Given that we have 0.081 moles of NH₃, we can plug in the values into the equation:
V = (0.081 mol)(0.082 L·atm/mol·K)(273.15 K) / (1 atm)
V = 1.8136 L
Therefore, the volume of ammonia produced in this reaction is 1.8136 liters.
When we have the moles of a gas produced in a reaction and know the conditions under which the gas is produced, we can use the ideal gas equation to calculate the volume of the gas. In this case, we had determined the moles of ammonia produced to be 0.081 mol and used STP conditions (1 atm pressure and 273.15 K temperature) to calculate the volume produced.
The ideal gas equation, PV = nRT, allows us to relate the pressure, volume, number of moles, ideal gas constant, and temperature of a gas. By rearranging the equation to solve for volume (V), we can plug in the known values and calculate the volume of the gas produced.
In this specific reaction involving Mg₃N₂ and ammonia production, we found that the volume of ammonia produced was 1.8136 liters. This calculation helps in understanding the stoichiometry of the reaction and determining the amount of product formed under given conditions.