Question

In the equation below

Answer 1

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH _3 }_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N _2 }_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H _2 O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

• The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

• The ratio between NH₃ and Cu is 2:3.
• The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

• The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH _3 } \cdot \frac{1\text{ mol N _2 }}{2\text{ mol NH _3 }} = 6.25\text{ mol N _2 }$

• The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH _3 } \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH _3 }} = 18.8\text{ mol Cu}$

• And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH _3 } \cdot \frac{3\text{ mol H _2 O}}{2\text{ mol NH _3 }} = 18.8\text{ mol H _2 O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.