Question

2Li+F2->2LiF
what mass of lithium was used if 87g f LiFis produced

Answer 1

To determine the mass of lithium used to produce 87 g of LiF, we can first use the molar mass of LiF and the stoichiometry of the reaction.

1. **Calculate the molar mass of LiF:**
- Molar mass of Li (Lithium) = 6.94 g/mol
- Molar mass of F (Fluorine) = 19.00 g/mol

\[
\text{Molar mass of LiF} = \text{Molar mass of Li} + \text{Molar mass of F} = 6.94 \, \text{g/mol} + 19.00 \, \text{g/mol} = 25.94 \, \text{g/mol}
\]

2. **Calculate the number of moles of LiF produced:**
\[
\text{Number of moles of LiF} = \frac{\text{mass of LiF}}{\text{molar mass of LiF}} = \frac{87 \, \text{g}}{25.94 \, \text{g/mol}} \approx 3.36 \, \text{moles of LiF}
\]

3. **Determine the number of moles of lithium used:**
From the balanced reaction \(2 \text{Li} + \text{F}_2 \rightarrow 2 \text{LiF}\), we see that 2 moles of Li produce 2 moles of LiF. Therefore, the number of moles of lithium used is the same as the number of moles of LiF produced.
\[
\text{Moles of Li used} = 3.36 \, \text{moles of LiF} = 3.36 \, \text{moles of Li}
\]

4. **Calculate the mass of lithium used:**
\[
\text{Mass of Li} = \text{moles of Li} \times \text{molar mass of Li} = 3.36 \, \text{moles} \times 6.94 \, \text{g/mol} \approx 23.29 \, \text{g}
\]

Therefore, the mass of lithium used to produce 87 g of LiF is approximately **23.29 g**.