`2NH_(3) to N_(2) + 3H_(2)`
If at the reaction, 18 mole of `H_(2)` is produced then find moles of
`NH_(3)` initially taken

Answer

To solve the problem, we will use stoichiometry based on the balanced chemical equation provided:

**Step 1: Write down the balanced chemical equation.**
The balanced equation is:
\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]

**Step 2: Identify the mole ratio from the balanced equation.**
From the equation, we can see that:
- 2 moles of NH₃ produce 3 moles of H₂.

**Step 3: Set up the relationship between NH₃ and H₂.**
From the mole ratio, we can express the amount of NH₃ needed to produce a certain amount of H₂:
- For every 3 moles of H₂ produced, 2 moles of NH₃ are consumed.

**Step 4: Determine how many moles of NH₃ are needed for 18 moles of H₂.**
We know that 18 moles of H₂ are produced. We can set up a proportion based on the mole ratio:
\[ \text{If } 3 \text{ moles of H}_2 \text{ require } 2 \text{ moles of NH}_3, \]
\[ \text{Then } 18 \text{ moles of H}_2 \text{ require } x \text{ moles of NH}_3. \]
Using the ratio:
\[ \frac{2 \text{ moles of NH}_3}{3 \text{ moles of H}_2} = \frac{x \text{ moles of NH}_3}{18 \text{ moles of H}_2} \]

**Step 5: Cross-multiply and solve for x.**
Cross-multiplying gives us:
\[ 2 \times 18 = 3 \times x \]
\[ 36 = 3x \]
Now, divide both sides by 3:
\[ x = \frac{36}{3} = 12 \]

**Step 6: Conclusion.**
Thus, the number of moles of NH₃ initially taken is **12 moles**.

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