Question

Ammonia gas can be produced by the reaction of nitrogen and hydrogen gases as shown in the following balanced equation:

N₂ (9) + 3H₂(g) → 2NH3 (9)
Determine the mass of ammonia that can be produced from 40.5 g of
N₂ (9) with excess
H₂ (9).
Mass=
9

Answer 1

Step-by-step explanation:

The first step is to determine the limiting reactant. In this case, we have 40.5 g of N₂ and excess H₂. We can use the molar mass of each substance to determine the number of moles of each reactant:

N₂: 40.5 g / 28.01 g/mol = 1.45 mol

H₂: 40.5 g / 2.016 g/mol = 20.1 mol

Since we have less moles of N₂ than H₂, N₂ is the limiting reactant. This means that the amount of ammonia produced will be determined by the amount of N₂.

We can use the balanced equation to determine the mass of ammonia produced:

1 mol N₂ → 2 mol NH₃

1.45 mol N₂ → 2.90 mol NH₃

2.90 mol NH₃ * 17.03 g/mol = 50.4 g NH₃

Therefore, 50.4 g of ammonia can be produced from 40.5 g of N₂ with excess H₂.

Answer 2

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To determine the mass of ammonia that can be produced from 40.5 g of N₂, we need to use stoichiometry and the balanced chemical equation provided.

From the balanced equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, we need to calculate the number of moles of N₂ in 40.5 g:

Number of moles of N₂ = mass / molar mass = 40.5 g / 28 g/mol = 1.4464 mol N₂

Using the mole ratio from the balanced equation, we can calculate the number of moles of NH₃ produced:

Number of moles of NH₃ = (1.4464 mol N₂) x (2 mol NH₃ / 1 mol N₂) = 2.8928 mol NH₃

Finally, we can use the molar mass of NH₃ to convert the number of moles to mass:

Mass of NH₃ = number of moles x molar mass = 2.8928 mol x 17 g/mol = 49.11 g

Therefore, the mass of ammonia that can be produced from 40.5 g of N₂ with excess H₂ is 49.11 g.