How many moles of Mg can reduce one mole of dil. `HNO_3` into `NH_(4)^(+)` ions ?

Answer

To solve the question of how many moles of magnesium (Mg) can reduce one mole of dilute HNO3 into NH4+ ions, we will follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
We need to establish the reaction that occurs when magnesium reacts with dilute nitric acid (HNO3). The balanced equation for this reaction is:
\[ 4 \text{Mg} + 10 \text{HNO}_3 \rightarrow 4 \text{Mg(NO}_3\text{)}_2 + \text{NH}_4\text{NO}_3 + 3 \text{H}_2\text{O} \]

2. **Identify the Moles of Reactants**:
From the balanced equation, we see that 4 moles of magnesium react with 10 moles of HNO3 to produce 1 mole of NH4+ ions.

3. **Determine the Ratio of Mg to HNO3**:
The ratio of magnesium to nitric acid from the balanced equation is:
\[ \frac{4 \text{ moles of Mg}}{10 \text{ moles of HNO}_3} \]
This simplifies to:
\[ \frac{2 \text{ moles of Mg}}{5 \text{ moles of HNO}_3} \]

4. **Calculate the Moles of Mg Required for 1 Mole of HNO3**:
To find out how many moles of magnesium are required to reduce 1 mole of HNO3 to NH4+, we can set up a proportion:
\[ \text{If } 10 \text{ moles of HNO}_3 \text{ require } 4 \text{ moles of Mg, then } 1 \text{ mole of HNO}_3 requires } x \text{ moles of Mg} \]
Using cross-multiplication:
\[ x = \frac{4 \text{ moles of Mg}}{10 \text{ moles of HNO}_3} \times 1 \text{ mole of HNO}_3 = 0.4 \text{ moles of Mg} \]

5. **Final Conclusion**:
Therefore, the number of moles of magnesium required to reduce one mole of dilute HNO3 into NH4+ ions is:
\[ \text{Answer: } 0.4 \text{ moles of Mg} \]