Question

When nitrogen gas and hydrogen gas react, the product is gaseous ammonia. balance the equation for this reaction. if you start with 0.9 mol of n2 and 0.9 mol of h2, how many moles of ammonia, nh3, will you produce and how many moles of the excess reactant remains? balanced equation:

a. 0.9 mol nh3 is produced and 0.3 mol of excess reactant remains
b. 0.9 mol nh3 is produced and 0.3 mol of excess reactant remains
c. 0.6 mol nh3 is produced and 0.3 mol of excess reactant remains
d. 1.8 mol nh3 is produced and 0.6 mol of excess reactant remains
e. 0.6 mol nh3 is produced and 0.6 mol of excess reactant remains

Answer 1

In this chemical reaction, 0.6 mol of ammonia (NH₃) is produced from the given amounts of nitrogen (N₂) and hydrogen (H₂), with 0.3 mol of excess nitrogen remaining.

Step-by-step explanation:

When nitrogen gas (N₂) and hydrogen gas (H₂) react, the product is gaseous ammonia (NH₃). The balanced equation for this reaction is: N₂(g) + 3H₂(g) → 2NH₃(g). The mole ratio of nitrogen to hydrogen to ammonia is 1:3:2.

Given that we are starting with 0.9 mol of N₂ and 0.9 mol of H₂, we first need to identify the limiting reactant. As per the balanced equation, 1 mole of N₂ requires 3 moles of H₂ for complete reaction. Therefore, to react with 0.9 mol of N₂, we would need 2.7 moles of H₂. Since we have only 0.9 mol of H₂, hydrogen is the limiting reactant.

Using the mole ratio of 3:2 for H₂ to NH₃, 0.9 mol of H₂ will produce (0.9 mol H₂) × (2 mol NH₃ / 3 mol H₂) = 0.6 mol NH₃. The excess reactant is N₂. Since the hydrogen is stoichiometrically limited, no H₂ remains; thus, 0.9 mol - 0.6 mol = 0.3 mol of N₂ will be in excess after the reaction.