Question

In the equation below

1. If a sample containing 12.5 moles of NH3 is reacted with excess CuO, how many moles of each product can be made?
N2=
Cu=
H20=

thanks!

Answer 1

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

Step-by-step 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.