`NH_(2)C"OO"NH_(4) overset(Delta)to 2NH_(3) + CO_(2)`
If 6 moles of `NH_(3)` is produced then find moles of `NH_(2) C"OO"NH_(4)` intially taken.

Answer

To solve the problem, we need to analyze the given chemical reaction and determine how many moles of the reactant \( NH_2COONH_4 \) were initially taken to produce 6 moles of \( NH_3 \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction provided is:
\[ NH_2COONH_4 \overset{\Delta}{\rightarrow} 2NH_3 + CO_2 \]
This equation indicates that 1 mole of \( NH_2COONH_4 \) produces 2 moles of \( NH_3 \).

2. **Identify the Relationship Between Reactants and Products:**
From the balanced equation, we see that:
- 1 mole of \( NH_2COONH_4 \) produces 2 moles of \( NH_3 \).
- Therefore, the ratio of \( NH_2COONH_4 \) to \( NH_3 \) is 1:2.

3. **Calculate Moles of \( NH_2COONH_4 \) Required for 6 Moles of \( NH_3 \):**
We need to find out how many moles of \( NH_2COONH_4 \) are required to produce 6 moles of \( NH_3 \).
Using the ratio derived from the balanced equation:
\[ \text{If } 2 \text{ moles of } NH_3 \text{ are produced from } 1 \text{ mole of } NH_2COONH_4, \]
then for 6 moles of \( NH_3 \):
\[ \text{Moles of } NH_2COONH_4 = \frac{6 \text{ moles of } NH_3}{2} = 3 \text{ moles of } NH_2COONH_4. \]

4. **Conclusion:**
Therefore, the initial amount of \( NH_2COONH_4 \) taken is **3 moles**.

### Final Answer:
The initial moles of \( NH_2COONH_4 \) taken is **3 moles**.

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