What volume of hydrogen, reacting with nitrogen, will produce 500 dm^3 of NH3 at STP?
To solve this problem, we will use the stoichiometry of the reaction and the ideal gas law.
The balanced chemical equation for the reaction between hydrogen gas (H2) and nitrogen gas (N2) to produce ammonia gas (NH3) is:
N2 (g) + 3H2 (g) → 2NH3 (g)
From the balanced equation, we can see that for every molecule of N2, 3 molecules of H2 are required, and 2 molecules of NH3 are produced. Since all the gases are at STP, we can use the molar volume of a gas at STP, which is 22.4 L/mol or 22.4 dm³/mol.
First, calculate the number of moles of NH3 produced:
volume of NH3 = 500 dm³;
molar volume of a gas at STP = 22.4 dm³/mol;
n(NH3) = volume of NH3 / molar volume = 500 dm³ / 22.4 dm³/mol = 22.32 mol of NH3
Now, use the stoichiometry to calculate the number of moles of H2:
From the balanced equation: 2 moles of NH3 are produced from 3 moles of H2
n(H2) = n(NH3) * (3 mol H2 / 2 mol NH3) = 22.32 mol * (3/2) = 33.48 mol of H2
Finally, find the volume of H2 at STP:
volume of H2 = n(H2) * molar volume = 33.48 mol * 22.4 dm³/mol = 749.95 dm³
So, approximately 750 dm³ of hydrogen gas is required to produce 500 dm³ of ammonia gas at STP.
To determine the volume of hydrogen gas needed to produce 500 dm^3 of NH3 at STP, we need to use the balanced chemical equation for the reaction between hydrogen and nitrogen, which is:
3H2 + N2 -> 2NH3
From the balanced equation, we can see that 3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia. Therefore, the molar ratio of hydrogen to ammonia is 3:2.
At STP, 1 mole of any gas occupies 22.4 dm^3. So, we can use this information to calculate the volume of hydrogen gas needed.
Step 1: Determine the number of moles of NH3 required.
Given volume of NH3 = 500 dm^3
At STP, 1 mole of NH3 occupies 22.4 dm^3.
Therefore, the number of moles of NH3 required = 500 dm^3 / 22.4 dm^3/mol = 22.32 mol
Step 2: Determine the number of moles of H2 required.
Since the molar ratio of H2 to NH3 is 3:2, we can set up a proportion:
3 mol H2 / 2 mol NH3 = x mol H2 / 22.32 mol NH3
Cross multiplying, we get:
3 mol H2 * 22.32 mol NH3 = 2 mol NH3 * x mol H2
x = (3 * 22.32) / 2
x = 33.48 mol H2
Step 3: Convert moles of H2 to volume.
At STP, 1 mole of any gas occupies 22.4 dm^3.
Therefore, the volume of H2 required = 33.48 mol * 22.4 dm^3/mol = 749.95 dm^3
Therefore, approximately 750 dm^3 of hydrogen gas will be needed to produce 500 dm^3 of NH3 at STP.
To determine the volume of hydrogen gas required to produce a given volume of ammonia gas, we need to use the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia:
N₂(g) + 3H₂(g) → 2NH₃(g)
According to the balanced equation, 3 moles of hydrogen gas react with 1 mole of nitrogen gas to produce 2 moles of ammonia gas.
Since we are given the volume at STP (Standard Temperature and Pressure), we can use the molar volume of a gas at STP, which is approximately 22.4 dm³/mol.
Now, let's calculate the number of moles of ammonia gas produced:
500 dm³ of NH₃ × (1 mol/22.4 dm³) = 22.32 mol
Since the stoichiometric ratio of H₂ to NH₃ is 3:2, we can calculate the number of moles of hydrogen gas required:
(22.32 mol NH₃) × (3 mol H₂/2 mol NH₃) = 33.48 mol H₂
Finally, we can convert the moles of hydrogen gas to volume using the molar volume at STP:
33.48 mol H₂ × (22.4 dm³/1 mol) = 749.95 dm³
Therefore, approximately 750 dm³ of hydrogen gas would be needed to produce 500 dm³ of ammonia gas at STP. Keep in mind that using approximate values for the molar volume and rounding errors during calculations may result in a slight discrepancy from the precise answer.