Atoms of three different elements are represented by O, h, and D. Which compound is left over when three molecules of OD and three molecules of hhD react to form OhD and ODD? I find this question so confusing.. can you please help?

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  1. Atoms of three different elements are represented by O, h, and D. Which compound is left over when three molecules of OD and three molecules of hhD react to form OhD and ODD?

I find this question so confusing.. can you please help?

### Conclusion

The leftover compound when three molecules of ☐Δ and three molecules of ☐☐Δ react to form ☐☐Δ and ☐ΔΔ is 2 ☐ ☐Δ.

---

#### Navigation Links:
- [Chapter 3, Problem 18ALQ](#)
- [Chapter 3, Problem 20ALQ](#)
Transcribed Image Text:### Conclusion The leftover compound when three molecules of ☐Δ and three molecules of ☐☐Δ react to form ☐☐Δ and ☐ΔΔ is 2 ☐ ☐Δ. --- #### Navigation Links: - [Chapter 3, Problem 18ALQ](#) - [Chapter 3, Problem 20ALQ](#)
### Answer to Problem 19ALQ

The compound that is leftover when three molecules of \(O△\) and three molecules of \(\square\square△\) react to form \(O\square△\) and \(O△△\) is \(2\square\square△\).

### Explanation of Solution

#### Explanation

**Given:**

The reaction that takes place is:
\[ 
3O△ + 3\square\square△ \rightarrow O\square△ + O△△.
\]

The number of molecules of each element on both sides is:

|   | Left side | Right side | Elements left |
|---|-----------|------------|---------------|
| **Circles**  | 3         | 2          | 1             |
| **Triangles**| 6         | 3          | 3             |
| **Squares**  | 6         | 1          | 5             |

On comparing the two sides, it is observed that the left side has 1 circle, 3 triangles, and 5 squares leftover. 

However, these leftover components are enough to make another molecule of \(\square\square△\) from the starting material. While it seems possible to form another molecule of \(△△\), this is not possible with the starting material because if another molecule of \(△△\) is made and the amounts of starting and ending products were compared, then the left side has 1 triangle and 5 squares more than the right side. 

1 triangle and 5 squares are not combined into any whole number multiple of the 1 triangle and 1 square needed to form \(△△\).
Transcribed Image Text:### Answer to Problem 19ALQ The compound that is leftover when three molecules of \(O△\) and three molecules of \(\square\square△\) react to form \(O\square△\) and \(O△△\) is \(2\square\square△\). ### Explanation of Solution #### Explanation **Given:** The reaction that takes place is: \[ 3O△ + 3\square\square△ \rightarrow O\square△ + O△△. \] The number of molecules of each element on both sides is: | | Left side | Right side | Elements left | |---|-----------|------------|---------------| | **Circles** | 3 | 2 | 1 | | **Triangles**| 6 | 3 | 3 | | **Squares** | 6 | 1 | 5 | On comparing the two sides, it is observed that the left side has 1 circle, 3 triangles, and 5 squares leftover. However, these leftover components are enough to make another molecule of \(\square\square△\) from the starting material. While it seems possible to form another molecule of \(△△\), this is not possible with the starting material because if another molecule of \(△△\) is made and the amounts of starting and ending products were compared, then the left side has 1 triangle and 5 squares more than the right side. 1 triangle and 5 squares are not combined into any whole number multiple of the 1 triangle and 1 square needed to form \(△△\).
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