Equal weights of Zn metal and iodine are mixed together and `I_(1)` is completley converted to `ZnI_(2)`. What fractionn by weight of original Zn remains unreacted? (Zn=65,I=127)

Options

A. 0.34
B. 0.74
C. 0.84
D. Unable to predict

Explanation

To solve the problem, we need to determine the fraction by weight of the original zinc (Zn) that remains unreacted after it reacts with iodine (I2) to form zinc iodide (ZnI2).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between zinc and iodine can be represented as:
\[ \text{Zn} + \text{I}_2 \rightarrow \text{ZnI}_2 \]

2. **Assume Equal Weights:**
Let the weight of zinc (Zn) and iodine (I2) taken initially be \( W \) grams each.

3. **Calculate Moles of Zinc and Iodine:**
- Moles of zinc:
\[ \text{Moles of Zn} = \frac{W}{65} \]
- Moles of iodine (I2):
\[ \text{Moles of I}_2 = \frac{W}{254} \]
(Note: The molar mass of I2 is \( 127 \times 2 = 254 \) g/mol)

4. **Identify the Limiting Reagent:**
The balanced equation shows that zinc and iodine react in a 1:1 ratio. Therefore, we compare the moles:
- Moles of Zn = \( \frac{W}{65} \)
- Moles of I2 = \( \frac{W}{254} \)

Since \( \frac{W}{254} < \frac{W}{65} \), iodine (I2) is the limiting reagent and will be completely consumed.

5. **Calculate Moles of Zinc Remaining:**
The moles of zinc that react will be equal to the moles of iodine:
\[ \text{Moles of Zn reacted} = \frac{W}{254} \]
Therefore, the moles of zinc remaining will be:
\[ \text{Moles of Zn remaining} = \frac{W}{65} - \frac{W}{254} \]

6. **Calculate the Weight of Zinc Remaining:**
The weight of zinc remaining can be calculated as:
\[ \text{Weight of Zn remaining} = \left( \frac{W}{65} - \frac{W}{254} \right) \times 65 \]

7. **Calculate the Fraction of Zinc Remaining:**
The fraction of the original weight of zinc that remains unreacted is given by:
\[ \text{Fraction of Zn remaining} = \frac{\text{Weight of Zn remaining}}{W} \]
Substituting the expression for weight of Zn remaining:
\[ \text{Fraction of Zn remaining} = \frac{\left( \frac{W}{65} - \frac{W}{254} \right) \times 65}{W} \]

8. **Simplify the Expression:**
Simplifying the fraction:
\[ = \frac{1}{65} - \frac{1}{254} \]
\[ = \frac{254 - 65}{65 \times 254} = \frac{189}{16510} \]
Now, calculating the fraction:
\[ = \frac{189}{16510} \approx 0.74 \]

### Final Answer:
The fraction by weight of the original zinc that remains unreacted is approximately **0.74**.