Problem 133 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The short answers are: a. \( 6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{Li}_{3}\mathrm{N}(s) \) b. \( 2\mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s) \)

Step 1: Identify the product

For the first reaction, Li (lithium) reacts with N (nitrogen) to form lithium nitride (Li3N), an ionic compound, where lithium has a charge of +1 and nitrogen a charge of -3. So the chemical equation becomes: \[ \mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s) \]

Step 2: Balance the equation

To balance the equation, we need to ensure there are the same number of Lithium and Nitrogen atoms on both sides. To do so, we can add coefficients in front of the elements and compounds. The balanced equation becomes: \[ 6\mathrm{Li}(s) + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{Li}_{3}\mathrm{N}(s) \] b.

Step 1: Identify the product

For the second reaction, Rb (rubidium) reacts with S (sulfur) to form rubidium sulfide (Rb2S), an ionic compound, where rubidium has a charge of +1 and sulfur a charge of -2. So the chemical equation becomes: \[ \mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s) \]

Step 2: Balance the equation

To balance the equation, we need to ensure there are the same number of Rubidium and Sulfur atoms on both sides of the equation. To do so, we can add coefficients in front of the elements and compounds. The balanced equation becomes: \[ 2\mathrm{Rb}(s) + \mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s) \]