To determine how many grams of ammonia (NH₃) will be formed from the complete reaction of 4.50 moles of hydrogen (H₂), we need to follow several steps, understanding the reaction stoichiometry and using the relationships between moles and mass. The balanced chemical equation for the reaction is given as:
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]
Here's a detailed step-by-step solution:
### Step 1: Understand the Stoichiometry of the Reaction
The balanced equation tells us that:
- 1 mole of [tex]\(N_2\)[/tex] reacts with 3 moles of [tex]\(H_2\)[/tex] to produce 2 moles of [tex]\(NH_3\)[/tex].
### Step 2: Determine the Mole Ratio
From the balanced equation, we see that 3 moles of [tex]\(H_2\)[/tex] produce 2 moles of [tex]\(NH_3\)[/tex]. Hence, the ratio of [tex]\(H_2\)[/tex] to [tex]\(NH_3\)[/tex] is:
[tex]\[ \frac{2 \ \text{moles} \ NH_3}{3 \ \text{moles} \ H_2} \][/tex]
### Step 3: Calculate the Moles of Ammonia ([tex]\(NH_3\)[/tex]) Produced
Given that we have 4.50 moles of [tex]\(H_2\)[/tex], we can calculate the moles of [tex]\(NH_3\)[/tex] produced using the mole ratio:
[tex]\[ \text{Moles of } NH_3 = 4.50 \ \text{moles} \ H_2 \times \frac{2 \ \text{moles} \ NH_3}{3 \ \text{moles} \ H_2} \][/tex]
[tex]\[ \text{Moles of } NH_3 = 4.50 \times \frac{2}{3} \][/tex]
[tex]\[ \text{Moles of } NH_3 = 3.00 \ \text{moles} \][/tex]
### Step 4: Calculate the Mass of Ammonia ([tex]\(NH_3\)[/tex]) Produced
To find the mass of ammonia produced, we use its molar mass. The molar mass of [tex]\(NH_3\)[/tex] is given as 17.03 g/mol. Therefore, the mass of [tex]\(NH_3\)[/tex] can be calculated as follows:
[tex]\[ \text{Mass of } NH_3 = \text{Moles of } NH_3 \times \text{Molar Mass of } NH_3 \][/tex]
[tex]\[ \text{Mass of } NH_3 = 3.00 \ \text{moles} \times 17.03 \ \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } NH_3 = 51.09 \ \text{grams} \][/tex]
### Conclusion
The complete reaction of 4.50 moles of hydrogen ([tex]\(H_2\)[/tex]) will produce 3.00 moles of ammonia ([tex]\(NH_3\)[/tex]), which is equivalent to 51.09 grams of ammonia.
