How many moles of ammonia gas can be formed from the complete reaction of 44.8 liters of nitrogen gas at standard temperature and pressure (STP), according to the balanced equation, N2 (g) + 3H2 (g) arrow 2NH3 (g)?
To determine the number of moles of ammonia gas formed, we need to use the given volume of nitrogen gas and the stoichiometry of the balanced equation.
First, let's convert the volume of nitrogen gas from liters to moles at STP. STP conditions are defined as 1 atmosphere (atm) of pressure and 273.15 Kelvin (K) temperature.
At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can calculate the number of moles of nitrogen gas using the following conversion:
moles of nitrogen gas = volume of nitrogen gas (liters) / 22.4 liters/mole
moles of nitrogen gas = 44.8 liters / 22.4 liters/mole
moles of nitrogen gas = 2 moles
According to the balanced equation, 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia gas (NH3). Therefore, the stoichiometric ratio is:
1 mole N2 : 3 moles H2 : 2 moles NH3
Since we have 2 moles of nitrogen gas, we can multiply the stoichiometric ratio to find the moles of ammonia gas formed:
moles of ammonia gas = 2 moles N2 * (2 moles NH3 / 1 mole N2)
moles of ammonia gas = 4 moles NH3
Therefore, 44.8 liters of nitrogen gas at STP can produce 4 moles of ammonia gas.