Answer: The amount of lithium reacted is 64.0215 grams.
Explanation:
To calculate the number of moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) ....(1)
Given mass of nitrogen = 86.1 g
Molar mass of nitrogen = 28 g/mol
Putting values in above equation, we get:
\(\text{Moles of nitrogen}=\frac{86.1g}{28g/mol}=3.075mol\)
For the reaction of nitrogen and lithium, the chemical equation follows:
\(3Li+N_2\rightarrow Li_3N_2\)
By Stoichiometry of the reaction:
1 mole of nitrogen reacts with 3 moles of lithium.
So, 3.075 moles of nitrogen will react with = \(\frac{3}{1}\times 3.075=9.225mol\) moles of lithium.
Now, calculating the mass of lithium by using equation 1, we get:
Molar mass of lithium = 6.94 g/mol
Moles of lithium = 9.225 mol
Putting values in equation 1, we get:
\(9.225mol=\frac{\text{Mass of lithium}}{6.94g/mol}\\\\\text{Mass of Lithium}=64.0215g\)
Hence, the amount of lithium reacted is 64.0215 grams.