Let's solve this problem step-by-step to determine the mass of lithium required to react completely with 58.9 mL of [tex]\( N_2 \)[/tex] gas at standard temperature and pressure (STP).
1. Convert the volume of [tex]\( N_2 \)[/tex] gas from mL to L:
Given:
[tex]\[
\text{Volume of } N_2 = 58.9 \text{ mL}
\][/tex]
Since 1 L = 1000 mL, we convert mL to L:
[tex]\[
\text{Volume of } N_2 = \frac{58.9 \text{ mL}}{1000} = 0.0589 \text{ L}
\][/tex]
2. Calculate the moles of [tex]\( N_2 \)[/tex] gas using the molar volume at STP:
At STP, 1 mole of any gas occupies 22.414 L. Therefore, the number of moles of [tex]\( N_2 \)[/tex] can be calculated using:
[tex]\[
\text{Moles of } N_2 = \frac{\text{Volume of } N_2}{\text{Molar volume at STP}} = \frac{0.0589 \text{ L}}{22.414 \text{ L/mol}} = 0.0026278218970286426 \text{ mol}
\][/tex]
3. Determine the moles of [tex]\( Li \)[/tex] required using the stoichiometric ratio from the balanced chemical equation:
The balanced chemical reaction is:
[tex]\[
6 Li (s) + N_2(g) \rightarrow 2 Li_3N(s)
\][/tex]
This tells us that 1 mole of [tex]\( N_2 \)[/tex] reacts with 6 moles of [tex]\( Li \)[/tex]. So, the moles of [tex]\( Li \)[/tex] required can be calculated as follows:
[tex]\[
\text{Moles of } Li = 0.0026278218970286426 \text{ mol} \times 6 = 0.015766931382171855 \text{ mol}
\][/tex]
4. Calculate the mass of [tex]\( Li \)[/tex] required:
The molar mass of lithium [tex]\( (Li) \)[/tex] is 6.94 g/mol. To find the mass of lithium required, we multiply the number of moles of [tex]\( Li \)[/tex] by its molar mass:
[tex]\[
\text{Mass of } Li = 0.015766931382171855 \text{ mol} \times 6.94 \text{ g/mol} = 0.10942250379227268 \text{ g}
\][/tex]
Therefore, the mass of lithium required to react completely with 58.9 mL of [tex]\( N_2 \)[/tex] gas at STP is approximately 0.109 grams.
