1N2 + 3H2 --> 2NH3

When 23.15 g of N, and 36.85 g of H2 react, what is the maximum amount of NH3 that can be produced?

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

the answer is definitely c.