Ammonia (NH) gas is produced via the exothermic reaction: N2g) + 3H2g) -> 2NH with 3.86 moles of nitrogen and 6.70 moles of hydrogen, what mass of ammonia concentration is produce?

Answer 1

Hi Laure! First, we need to consider our limiting reagent. Let us determine the mols of NH3 produced from EACH reagent:

From N2, there are 3.86 mols of N2. Since 1 mol of N2 produces 2 mols of NH3 from the reaction, we can say:

3.86 mols N2 * 2mols NH3/1 mol N2 = 7.72 mols NH3

From H2, there are 6.70 mols H2. Since 3 mols of H2 produce 2 mols of NH3 from the reaction, we can say:

6.40 mols H2 * 2 mols NH3/ 3 mols H2 = 4.27 mols NH3

Since the amount of H2 produces LESS NH3 when it reacts completely, then 4.27 mols of NH3 are produced. Since we want a mass of NH3, we can convert to grams using the molecular weight of NH3:

4.27 mols NH3 * 17.031 g NH3/mol = 72.7 g NH3

I hope this helps!

Answer 2

The problem gives us the molar amounts of our reactants (Nitrogen and Hydrogen). In order to move on, we must first determine which one is the limiting reagent. To do this, we look at our balanced chemical equation:

N₂ + 3H₂ --> 2NH₃

The coefficients in front of each component represents the molar equivalents that reaction uses or creates. For every 1 mole of N₂, we should have 3 moles of H₂. Looking at the problem, we notice that we have 3.86 moles of N₂. By our equation's standards, then, we should have 11.58 moles of H₂, but we only have 6.70 moles. Therefore, H₂ is our limiting reagent! So we will use 6.70 as our 3-molar-equivalent measure for H₂.

6.70 moles = 3 molar equivalents

2.23 moles = 1 molar equivalent

4.46 moles = 2 molar equivalents

Based on the limiting reagent, we will create 2 molar equivalents of NH₃, or 4.46 moles. We can convert moles of NH₃ to grams of NH₃ using its molar mass, 17.031 g/mol.

4.46 moles * 17.031 g/mol = 75.96 g NH₃