`XeF_(6)` fluorinates `I_(2)` to `IF_(7)` and liberates Xenon (g). `210 mmol` of `XeF_(6)` can yield a maximum of `_ mmol` of `IF_(7)` :

Options

A. 420
B. 180
C. 210
D. 245

Explanation

To solve the problem, we need to determine how many millimoles of \( IF_7 \) can be produced from \( 210 \, \text{mmol} \) of \( XeF_6 \). We will do this by first writing the balanced chemical equation for the reaction and then using stoichiometry to find the yield of \( IF_7 \).

### Step 1: Write the balanced chemical equation

The reaction is given as:
\[ XeF_6 + I_2 \rightarrow IF_7 + Xe \]

From the video transcript, we find the balanced equation:
\[ 7 \, XeF_6 + 3 \, I_2 \rightarrow 6 \, IF_7 + 7 \, Xe \]

### Step 2: Determine the mole ratio

From the balanced equation, we can see the mole ratio between \( XeF_6 \) and \( IF_7 \):
- 7 moles of \( XeF_6 \) produce 6 moles of \( IF_7 \).

### Step 3: Calculate the moles of \( IF_7 \) produced from \( 210 \, \text{mmol} \) of \( XeF_6 \)

Using the mole ratio, we can set up a proportion to find out how many millimoles of \( IF_7 \) can be produced from \( 210 \, \text{mmol} \) of \( XeF_6 \):

\[ \text{Moles of } IF_7 = \left( \frac{6 \, \text{moles of } IF_7}{7 \, \text{moles of } XeF_6} \right) \times 210 \, \text{mmol} \]

### Step 4: Perform the calculation

Now we can calculate the number of millimoles of \( IF_7 \):
\[ \text{Moles of } IF_7 = \left( \frac{6}{7} \right) \times 210 = 180 \, \text{mmol} \]

### Final Answer

Thus, the maximum number of millimoles of \( IF_7 \) that can be produced from \( 210 \, \text{mmol} \) of \( XeF_6 \) is:
\[ \boxed{180 \, \text{mmol}} \]

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