What volume of ammonia gas will be produced when 1.44 L of nitrogen react completely in the following equation?
N2 + 3H2 = 2NH3
To determine the volume of ammonia gas produced, we need to use the stoichiometry of the equation, which relates the number of moles of reactants and products.
Given that 1.44 L of nitrogen reacts completely, we first need to convert this volume to moles. To do this, we need to know the conditions (temperature and pressure) under which the nitrogen gas is present. This information is crucial because it allows us to use the ideal gas law to calculate the number of moles.
Let's assume the nitrogen gas is at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere (atm) of pressure.
At STP, one mole of any ideal gas occupies 22.4 liters. Therefore, we can calculate the number of moles of nitrogen gas as follows:
Number of moles of N2 = (Volume of N2 gas) / (Molar volume at STP)
= 1.44 L / 22.4 L/mol
= 0.0643 moles of N2
From the balanced equation, we can see that the stoichiometric ratio between nitrogen and ammonia is 1:2. This means that for every mole of nitrogen gas that reacts, two moles of ammonia gas are produced.
Using this stoichiometric ratio, we can calculate the number of moles of ammonia produced:
Number of moles of NH3 = (Number of moles of N2) x (2 moles of NH3 / 1 mole of N2)
= 0.0643 moles of N2 x (2 moles of NH3 / 1 mole of N2)
= 0.1286 moles of NH3
Finally, to convert the moles of ammonia to volume, we can use the molar volume at STP:
Volume of NH3 gas = (Number of moles of NH3) x (Molar volume at STP)
= 0.1286 moles of NH3 x 22.4 L/mol
= 2.8814 L of NH3 gas
Therefore, when 1.44 L of nitrogen gas reacts completely, it produces 2.8814 L of ammonia gas.