`4Zn+10HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O`. In this reaction one mole of `HNO_(3)` is reduced by

Options

A. 32g Zn
B. 64g Zn
C. 128g Zn
D. 256g Zn

Answer

To determine how many grams of zinc (Zn) are required to reduce one mole of nitric acid (HNO₃) in the given reaction, we can follow these steps:

### Step 1: Analyze the Reaction
The balanced chemical reaction is:
\[ 4 \text{Zn} + 10 \text{HNO}_3 \rightarrow 4 \text{Zn(NO}_3)_2 + \text{NH}_4\text{NO}_3 + 3 \text{H}_2\text{O} \]

From this equation, we can see that 10 moles of HNO₃ react with 4 moles of Zn.

### Step 2: Determine the Zinc to HNO₃ Ratio
From the balanced equation:
- 10 moles of HNO₃ require 4 moles of Zn.
- Therefore, 1 mole of HNO₃ will require \( \frac{4}{10} = 0.4 \) moles of Zn.

### Step 3: Calculate the Molar Mass of Zinc
The molar mass of zinc (Zn) is approximately 65.38 g/mol.

### Step 4: Calculate the Mass of Zinc Required
To find the mass of zinc required for 0.4 moles, we use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
Substituting the values:
\[ \text{Mass of Zn} = 0.4 \, \text{moles} \times 65.38 \, \text{g/mol} \]
\[ \text{Mass of Zn} = 26.152 \, \text{g} \]

### Final Answer
Therefore, 26.152 grams of zinc are required to reduce one mole of HNO₃.

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