Question

__NG3 + ___CuO -> __N2 +__Cu + H2O

Answer 1

To balance the chemical equation involving the reaction of a nitrogen gas compound (possibly representing an amine or another compound containing nitrogen) with copper(II) oxide (CuO) to produce nitrogen gas (N2), copper metal (Cu), and water (H2O), we can use the following general steps:

1. **Identify the compounds and elements**: The unbalanced equation appears to involve nitrogen (N), copper (Cu), oxygen (O), and hydrogen (H).

2. **Write the unbalanced equation**:
\[
\text{NG}_3 + \text{CuO} \rightarrow \text{N}_2 + \text{Cu} + \text{H}_2\text{O}
\]

3. **Determine the coefficients**: Let's assume NG3 stands for \( \text{N}_2\text{H}_6 \) (which is an example of a nitrogen-hydrogen compound, such as hydrazine). The equation would be balanced as follows:

4. **Balancing the equation**:
- The left side contains nitrogen (N), copper (Cu), and oxygen (O).
- We assign coefficients to balance the nitrogen, copper, and oxygen atoms.

Assuming a nitrogen compound like hydrazine \( \text{N}_2\text{H}_6 \) might react:

\[
2 \text{N}_2\text{H}_6 + 6 \text{CuO} \rightarrow 2 \text{N}_2 + 6 \text{Cu} + 6 \text{H}_2\text{O}
\]

This gives us:

- **Nitrogen**: 4 on both sides (2 from \( \text{N}_2\text{H}_6 \) and 2 from \( \text{N}_2 \)).
- **Copper**: 6 on both sides (from \(6 \text{CuO}\) to produce \(6 \text{Cu}\)).
- **Oxygen**: 6 on both sides (from \(6 \text{CuO}\) to produce \(6 H_2O\)).
- **Hydrogen**: 12 on both sides (from \(2 \text{N}_2\text{H}_6\) to produce \(6 \text{H}_2\text{O}\)).

Thus, the balanced equation would look like:

\(
2 \text{N}_2\text{H}_6 + 6 \text{CuO} \rightarrow 2 \text{N}_2 + 6 \text{Cu} + 6 \text{H}_2\text{O}
\)

If "NG3" represents another specific nitrogen compound, the coefficients may vary, but this general approach outlines how to balance such reactions.