To determine how many moles of lithium (Li) are produced when 325 grams of barium (Ba) react completely, we can follow these steps:
### Step 1: Write down the given information
- Mass of barium (Ba): [tex]\(325 \text{ grams}\)[/tex]
- Molar mass of barium (Ba): [tex]\(137.327 \text{ g/mol}\)[/tex]
### Step 2: Calculate the number of moles of barium (Ba)
The number of moles is calculated using the formula:
[tex]\[
\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}
\][/tex]
[tex]\[
\text{Moles of barium} = \frac{325 \text{ g}}{137.327 \text{ g/mol}} \approx 2.37 \text{ moles}
\][/tex]
### Step 3: Use the stoichiometry of the balanced chemical equation
The balanced equation is:
[tex]\[
2 \text{LiBr} + \text{Ba} \rightarrow \text{BaBr}_2 + 2 \text{Li}
\][/tex]
From the balanced reaction, we see that 1 mole of barium (Ba) produces 2 moles of lithium (Li).
### Step 4: Calculate the moles of lithium (Li) produced
Since 1 mole of barium produces 2 moles of lithium, we can double the moles of barium to find the moles of lithium:
[tex]\[
\text{Moles of lithium} = 2 \times \text{Moles of barium} = 2 \times 2.37 \approx 4.73 \text{ moles}
\][/tex]
### Step 5: Conclusion
Therefore, the number of moles of lithium produced is approximately 4.73 moles. The correct answer is:
C. [tex]\(\quad 4.73 \text{ moles}\)[/tex]
