Answer:
The amount of ammonia (NH3) produced would be 30.26 grams. The limiting reactant in this reaction is nitrogen gas (N2), and the excess reactant is hydrogen gas (H2).
Step-by-step explanation:
To determine the limiting reactant and excess reactant, we need to compare the stoichiometry of the balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3).
The balanced chemical equation for the reaction is:
N2 + 3H2 → 2NH3
From the equation, we can see that the mole ratio of N2 to NH3 is 1:2, and the mole ratio of H2 to NH3 is 3:2.
Given that we have 25 grams of N2 and 25 grams of H2, we can calculate the number of moles of each:
Molar mass of N2 = 28 g/mol
Molar mass of H2 = 2 g/mol
Moles of N2 = 25 g / 28 g/mol = 0.89 mol
Moles of H2 = 25 g / 2 g/mol = 12.5 mol
According to the mole ratios in the balanced equation, we can see that 0.89 mol of N2 would react with 0.89 x 2 = 1.78 mol of NH3, and 12.5 mol of H2 would react with 12.5 x 2/3 = 8.33 mol of NH3.
Since the mole ratio of N2 to NH3 is 1:2, and we have only 0.89 mol of N2, N2 is the limiting reactant as it would produce the least amount of NH3. The excess reactant is H2, as it would have some amount left over after the reaction is complete.
To calculate the mass of NH3 produced, we can use the mole ratio of NH3 to N2, which is 2:1:
Molar mass of NH3 = 17 g/mol
Moles of NH3 = 0.89 mol of N2 x 2/1 = 1.78 mol of NH3
Mass of NH3 = 1.78 mol x 17 g/mol = 30.26 g of NH3