When 6 liter of nitrogen gas reacts with 18
liters of hydrogen gas at constant temperature
and pressure, how many liters of ammonia gas
will be produced?
N2 + 3H2 ==> 2NH3
When gases are involved one may use a shortcut and use L as if they were mols.
This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants.
I do these the long way.
How much NH3 can be produced from 6 L N2 and all of the H2 needed. That is
6L x (2 mol NH3/1 mol H2) = 12 L NH3.
How much NH3 can be produced from 18 L H2 and all of the N2 needed. That is
18 x (2 mols NH3/3 mols H2) = 12 L NH3.
The numbers are the same so you will have 18 L NH3 produced.
To determine the amount of ammonia gas produced, we need to first write a balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia gas (NH3):
N2 + 3H2 → 2NH3
According to the balanced chemical equation, one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas.
Given that the volume of nitrogen gas is 6 liters and the volume of hydrogen gas is 18 liters, we can use the concept of stoichiometry to determine the volume of ammonia gas produced.
Step 1: Convert the given volumes to moles using the ideal gas law.
1. Calculate the number of moles of nitrogen gas (N2):
moles of N2 = volume of N2 / molar volume of N2
Assuming the temperature and pressure are constant, we can use the molar volume of an ideal gas at standard temperature and pressure (STP):
molar volume of N2 = 22.4 L/mol
moles of N2 = 6 L / 22.4 L/mol = 0.268 moles of N2
2. Calculate the number of moles of hydrogen gas (H2):
moles of H2 = volume of H2 / molar volume of H2
Similarly, using the molar volume of an ideal gas at STP:
molar volume of H2 = 22.4 L/mol
moles of H2 = 18 L / 22.4 L/mol = 0.804 moles of H2
Step 2: Use the mole ratios from the balanced chemical equation to determine the moles of NH3 formed.
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Therefore, the moles of NH3 can be calculated as follows:
moles of NH3 = (moles of N2) x (moles of NH3 / moles of N2)
= 0.268 moles of N2 x (2 moles of NH3 / 1 mole of N2)
= 0.536 moles of NH3
Step 3: Convert the moles of NH3 to volume using the ideal gas law.
Using the molar volume of an ideal gas at STP:
molar volume of NH3 = 22.4 L/mol
volume of NH3 = moles of NH3 x molar volume of NH3
= 0.536 moles of NH3 x 22.4 L/mol
= 12.0064 L
Rounded to the appropriate number of significant figures, the volume of ammonia gas produced is approximately 12 liters.
To find the number of liters of ammonia gas produced, we need to determine the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia. The balanced equation is as follows:
N₂ + 3H₂ → 2NH₃
From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Next, we need to calculate the number of moles of nitrogen and hydrogen given the volume of gas provided. The molar volume of a gas at standard conditions (0°C and 1 atm) is 22.4 liters/mol. Therefore, the number of moles of nitrogen is:
moles of N₂ = volume of N₂ gas / molar volume = 6 liters / 22.4 liters/mol = 0.268 moles
Similarly, the number of moles of hydrogen is:
moles of H₂ = volume of H₂ gas / molar volume = 18 liters / 22.4 liters/mol = 0.804 moles
According to the stoichiometry of the balanced equation, the limiting reactant is the one with a lower number of moles. In this case, nitrogen is the limiting reactant since it has fewer moles compared to hydrogen. It means that nitrogen will be completely consumed in the reaction, and some excess hydrogen may be left.
Since 1 mole of nitrogen reacts with 2 moles of ammonia, we can determine the number of moles of ammonia produced using the stoichiometry:
moles of NH₃ = moles of N₂ × (2 moles of NH₃ / 1 mole of N₂) = 0.268 moles × (2/1) = 0.536 moles
Finally, to find the volume of ammonia gas produced, we can use the molar volume of the gas at standard conditions:
volume of NH₃ gas = moles of NH₃ × molar volume = 0.536 moles × 22.4 liters/mol = 12.0 liters
Therefore, when 6 L of nitrogen gas reacts with 18 L of hydrogen gas, 12 L of ammonia gas will be produced.