To determine the maximum mass of NH3 that can be produced, we need to balance the chemical equation first. The balanced equation for the reaction between N2 and H2 to form NH3 is:
N2 + 3H2 → 2NH3
Now, let's calculate the maximum mass of NH3 that can be produced using the given masses of N2 and H2.
Molar mass of N2 = 28.02 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of NH3 = 17.03 g/mol
To find the limiting reactant, we need to compare the number of moles of N2 and H2. We'll use the molar masses to convert the given masses into moles:
Moles of N2 = 1.0 g / 28.02 g/mol ≈ 0.036 mol
Moles of H2 = 1.0 g / 2.02 g/mol ≈ 0.495 mol
According to the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. This means that 1 mole of N2 will produce 2 moles of NH3.
Since the ratio between N2 and H2 is 1:3, we need 3 moles of H2 for every 1 mole of N2. Therefore, the maximum moles of NH3 that can be produced will be twice the number of moles of N2:
Maximum moles of NH3 = 2 × moles of N2 ≈ 2 × 0.036 mol = 0.072 mol
Finally, we can calculate the maximum mass of NH3 using the molar mass of NH3:
Maximum mass of NH3 = Maximum moles of NH3 × Molar mass of NH3
= 0.072 mol × 17.03 g/mol
≈ 1.2276 g
Therefore, the maximum mass of NH3 that can be produced is approximately 1.2276 grams.