Question

Initially there are 0.220 mol N₂ and 0.701 mol H₂ present. If a reaction occurs between N₂ and H₂ to form NH₃, how many moles of NH₃ can be produced, assuming complete reaction?

a. 0.220 mol
b. 0.701 mol
c. 0.921 mol
d. 0.481 mol

Answer 1

The balanced equation for the reaction is N₂ + 3H₂ → 2NH₃. The moles of NH₃ produced can be calculated using the stoichiometric ratio, and the limiting reactant is N₂.

Step-by-step explanation:

The balanced equation for the reaction is: N₂ + 3H₂ → 2NH₃. From the equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Given that there are initially 0.220 mol of N₂ and 0.701 mol of H₂, we need to determine the limiting reactant.

To do this, we compare the moles of N₂ and H₂ to the stoichiometric ratio.

For N₂: H₂, the ratio is 1:3. Since there are more moles of H₂ than the stoichiometric ratio requires, H₂ is in excess. Therefore, the limiting reactant is N₂.

Using the stoichiometric ratio, we can calculate the moles of NH₃ produced. Given 0.220 mol of N₂, we have:

(0.220 mol N₂) × (2 mol NH₃ / 1 mol N₂) = 0.440 mol NH₃

Therefore, the correct answer is 0.440 mol NH₃.