Answer:
56.67 g of NH₃.
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N₂ + 3H₂ —> 2NH₃
Next, we shall determine the masses of N₂ and H₂ that reacted and the mass of NH₃ produced from the balanced equation. This is illustrated below:
Molar mass of N₂ = 14 × 2 = 28 g/mol
Mass of N₂ from the balanced equation = 1 × 28 = 28 g
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H₂ from the balanced equation = 3 × 2 = 6 g
Molar mass of NH₃ = 14 + (3×1) = 14 + 3 = 17 g/mol
Mass of NH₃ from the balanced equation = 2 × 17 = 34 g
SUMMARY:
From the balanced equation above,
28 g of N₂ reacted with 6 g of H₂ to produce 34 g of NH₃
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
28 g of N₂ reacted with 6 g of H₂.
Therefore, 80 g of N₂ will react with = (80 × 6)/28 = 17.14 g of H₂.
We can see from the calculation made above that a higher mass (i.e 17.14 g) of H₂ than what was given (i.e 10 g) is required to react completely with 80 g of N₂. Thus, H₂ is the limiting reactant and N₂ is the excess reactant.
Finally, we shall determine the maximum mass of ammonia (NH₃) produced from the reaction.
To obtain the maximum mass of NH₃, the limiting reactant will be used because all of it is consumed in the reaction.
The limiting reactant is H₂ and the maximum mass of NH₃ can be obtained as follow:
From the balanced equation above,
6 g of H₂ reacted to produce 34 g of NH₃.
Therefore, 10 g of H₂ will react to produce = (10 × 34)/6 = 56.67 g of NH₃.
Therefore, 56.67 g of NH₃ is produced from the reaction.