Balance this chemical equation and assign it a reaction type: \(\mathrm{Li}(\mathrm{s})+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\)

We will start by counting the number of atoms for each element on both sides of the equation: On the left-hand side (reactants): 1 Li and 2 N atoms On the right-hand side (products): 3 Li and 1 N atom

We'll begin by balancing the nitrogen atoms. To balance the nitrogen atoms, we need to multiply the nitrogen molecule on the left side by 1/2: \(1 \, \mathrm{Li}(\mathrm{s}) + \frac{1}{2} \, \mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3} \mathrm{N}(s)\) However, it is not allowed to have fractions in a balanced equation. Therefore, we will multiply the entire equation by 2 to eliminate the fraction: \(2 \, \mathrm{Li}(\mathrm{s}) + \mathrm{N}_{2}(g) \rightarrow 2 \, \mathrm{Li}_{3} \mathrm{N}(s)\) Now the nitrogen atoms are balanced: On the left-hand side: 2 N atoms On the right-hand side: 2 N atoms Next, we will balance the lithium atoms. We already have 2 Li atoms on the left side, and with the multiplication by 2 in the previous step, we have 6 Li atoms on the right side. Therefore, we need to change the coefficient for Li on the left side to 6: \(6 \, \mathrm{Li}(\mathrm{s}) + \mathrm{N}_{2}(g) \rightarrow 2 \, \mathrm{Li}_{3} \mathrm{N}(s)\) Now the lithium atoms are balanced: On the left-hand side: 6 Li atoms On the right-hand side: 6 Li atoms So the balanced equation is: \(6 \, \mathrm{Li}(\mathrm{s}) + \mathrm{N}_{2}(g) \rightarrow 2 \, \mathrm{Li}_{3} \mathrm{N}(s)\)

This chemical reaction involves a metal (lithium) reacting with a non-metal (nitrogen). When a metal reacts with a non-metal to form a compound, the reaction is typically a synthesis reaction (also called a combination reaction). In conclusion, the balanced chemical equation is: \(6 \, \mathrm{Li}(\mathrm{s}) + \mathrm{N}_{2}(g) \rightarrow 2 \, \mathrm{Li}_{3} \mathrm{N}(s)\) and the reaction type is a synthesis (combination) reaction.