Complete and balance the equations for the following reactions. a. \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow\) b. \(\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow\)

First, let's write down the given chemical equation: \[\mathrm{Li}(s) + \mathrm{N}_2(g) \rightarrow\] To balance this equation, we need to ensure that the number of Lithium (Li) and Nitrogen (N) atoms are equal on both sides. Since there are two Nitrogen atoms in the Nitrogen gas molecule, we should form a compound with an even number of Nitrogen atoms. The most common compound formed between Lithium and Nitrogen is Lithium nitride (Li3N). Now, let's rewrite the equation: \[\mathrm{Li}(s) + \mathrm{N}_2(g) \rightarrow \mathrm{Li}_3\mathrm{N}(s)\] Now, we need to balance the Lithium atoms. There are 3 Lithium atoms in the Lithium nitride, so we need to add a coefficient of 3 in front of the Lithium atoms on the left side: \[\mathrm{3}\mathrm{Li}(s) + \mathrm{N}_2(g) \rightarrow \mathrm{Li}_3\mathrm{N}(s)\] The equation is now balanced.

Let's write down the given chemical equation: \[\mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow\] To balance this equation, we need to ensure that the number of Rubidium (Rb) and Sulfur (S) atoms are equal on both sides. The most common compound formed between Rubidium and Sulfur is Rubidium sulfide (Rb2S). Now, let's rewrite the equation: \[\mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow \mathrm{Rb}_2\mathrm{S}(s)\] Now, we need to balance the Rubidium atoms. There are 2 Rubidium atoms in the Rubidium sulfide, so we need to add a coefficient of 2 in front of the Rubidium atoms on the left side: \[\mathrm{2}\mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow \mathrm{Rb}_2\mathrm{S}(s)\] The equation is now balanced.