Flourine reacts with uranium to produce uranium hexaflouride ,
`UF_(6)`, as represented by this equation
`U(s) + 3F_(2)(g) to UF_(6)(g)`
How many fluorine molecules are required to produce 7.04 mg of uranium hexafluoride ,
`UF_(6)` , from an excess of uranium? The molar mass of `UF_(6)` is 352 gm/mol `(N_(A)=6xx10^(23))`

Answer

To solve the problem of how many fluorine molecules are required to produce 7.04 mg of uranium hexafluoride (UF₆), we will follow these steps:

### Step 1: Convert mass from milligrams to grams
Given mass of UF₆ = 7.04 mg

To convert milligrams to grams:
\[ \text{Mass in grams} = \frac{7.04 \text{ mg}}{1000} = 0.00704 \text{ g} \]

### Step 2: Calculate moles of UF₆
Using the molar mass of UF₆, which is 352 g/mol, we can calculate the number of moles of UF₆ using the formula:
\[ \text{Moles of UF₆} = \frac{\text{Mass of UF₆}}{\text{Molar mass of UF₆}} = \frac{0.00704 \text{ g}}{352 \text{ g/mol}} \]
\[ \text{Moles of UF₆} = 2 \times 10^{-5} \text{ mol} \]

### Step 3: Calculate the number of molecules of UF₆
To find the number of molecules, we use Avogadro's number (\(N_A = 6 \times 10^{23}\) molecules/mol):
\[ \text{Number of molecules of UF₆} = \text{Moles of UF₆} \times N_A = 2 \times 10^{-5} \text{ mol} \times 6 \times 10^{23} \text{ molecules/mol} \]
\[ \text{Number of molecules of UF₆} = 1.2 \times 10^{19} \text{ molecules} \]

### Step 4: Determine the number of fluorine molecules required
From the balanced chemical equation:
\[ U(s) + 3F_2(g) \rightarrow UF_6(g) \]
It shows that 1 mole of UF₆ requires 3 moles of F₂. Therefore, the number of moles of F₂ required is:
\[ \text{Moles of F₂} = 3 \times \text{Moles of UF₆} = 3 \times 2 \times 10^{-5} \text{ mol} = 6 \times 10^{-5} \text{ mol} \]

### Step 5: Calculate the number of molecules of F₂
Using Avogadro's number again:
\[ \text{Number of molecules of F₂} = \text{Moles of F₂} \times N_A = 6 \times 10^{-5} \text{ mol} \times 6 \times 10^{23} \text{ molecules/mol} \]
\[ \text{Number of molecules of F₂} = 3.6 \times 10^{19} \text{ molecules} \]

### Final Answer
The number of fluorine molecules required to produce 7.04 mg of uranium hexafluoride (UF₆) is:
\[ \boxed{3.6 \times 10^{19}} \text{ molecules} \]