What is the volume of NH3 produced in the following reaction when 3.0L of N2 reacts with 4.0L of H2?

Answer 1

N₂ + 3 H₂ ------> 2 NH₃

1 mole of N₂ produces 2 moles of NH₃; 1 mole of H₂ produces 2/3 mole of NH₃ ------> since 1 mole of H₂ produces less NH₃ than 1 mole of N₂, H₂ is the limiting reagent

4.0 L H₂ x (2 L NH₃ / 3 L H₂) = 2.67 L NH₃ produced

Answer 2

To solve this problem, we first need to balance the chemical equation:

N2+3H2 ---) 2NH3

From the balanced equation we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Next, we need to determine the limiting reactant, which is the reactant that will be completely consumed, thus limiting the amount of product formed. To do this, we compare the amounts of N2 and H2 provided:

1. N2: 3.0 L
2. H2: 4.0 L

According to the stoichiometry of the reaction, 1 mole of N2 requires 3 moles of H2. Therefore, to react completely, 3.0 L of N2 would require (3.0 L / 3) = 1.0 L of H2.

Since we have more than enough H2 (4.0 L), N2 is the limiting reactant.

Now, we'll use the stoichiometry of the reaction to determine the volume of NH3 produced. From the balanced equation, we know that 1 mole of N2 produces 2 moles of NH3.

Given that 3.0 L of N2 is consumed, and using the stoichiometry ratio of 1 mole N2 to 2 moles NH3, we find that:

Volume of NH3 produced = (3.0 L N2) × (2 mol NH3 / 1 mol N2) = 6.0 L

So, the volume of NH3 produced in the reaction is 6.0 liters.

This video may help explain limiting and excess reagents better: https://youtu.be/TiHnwU22LBg