Answer:
Step-by-step explanation:
To calculate the amount of NH₃(g) formed, we need to determine the limiting reactant in the given reaction. The limiting reactant is the reactant that is completely consumed, thereby limiting the amount of product that can be formed.
Given:
Mass of N₂(g) = 50 kg
Mass of H₂(g) = 10 kg
To find the limiting reactant, we need to compare the moles of N₂(g) and H₂(g) present.
The molar mass of N₂ is 28 g/mol, so the number of moles of N₂(g) can be calculated as:
Moles of N₂(g) = Mass of N₂(g) / Molar mass of N₂
= 50 kg / 28 g/mol
= 1785.71 mol
The molar mass of H₂ is 2 g/mol, so the number of moles of H₂(g) can be calculated as:
Moles of H₂(g) = Mass of H₂(g) / Molar mass of H₂
= 10 kg / 2 g/mol
= 5000 mol
According to the balanced equation, the stoichiometric ratio of N₂ to NH₃ is 1:2, and the stoichiometric ratio of H₂ to NH₃ is 3:2. This means that for every 1 mole of N₂, 2 moles of NH₃ are produced, and for every 3 moles of H₂, 2 moles of NH₃ are produced.
Comparing the moles of N₂ and H₂ to the stoichiometric ratios, we find that there is an excess of H₂. Therefore, H₂ is the limiting reactant.
To calculate the amount of NH₃(g) formed, we need to use the stoichiometric ratio between H₂ and NH₃. From the balanced equation, we know that for every 3 moles of H₂, 2 moles of NH₃ are produced.
Moles of NH₃(g) = (Moles of H₂(g) / 3) * 2
= (5000 mol / 3) * 2
= 3333.33 mol
Now, we can calculate the mass of NH₃(g) formed using the molar mass of NH₃, which is 17 g/mol.
Mass of NH₃(g) = Moles of NH₃(g) * Molar mass of NH₃
= 3333.33 mol * 17 g/mol
= 56666.67 g
= 56.67 kg
Therefore, the amount of NH₃(g) formed is approximately 56.67 kg.
The correct answer is A. 56.67 kg.