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Chapter 7: Problem 143

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

Short Answer

Expert verified
The short answers for the balanced equations are: a. \(6\operatorname{Li}(s)+\mathrm{N}_{2}(g) \rightarrow 2\operatorname{Li}_{3}\operatorname{N}(s)\) b. \(2\operatorname{Rb}(s)+\mathrm{S}(s) \rightarrow \operatorname{Rb}_{2}\operatorname{S}(s)\)

Step by step solution

01

(a) Identify the product and its stoichiometry

For the reaction between lithium (Li) and nitrogen (N), the product will be a lithium nitride compound. Since lithium has a +1 charge and nitrogen has a -3 charge, the compound formed will be Li3N, with a balanced ratio of charges.
02

(a) Write the unbalanced equation

Now, we can write the unbalanced equation for the reaction: \( \operatorname{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \operatorname{Li}_{3}\operatorname{N}(s) \)
03

(a) Balance the equation

To balance the equation, we need to make sure there are equal numbers of each element on both sides of the reaction. We can use the method of coefficients to achieve this: \(6 \operatorname{Li}(s)+\mathrm{N}_{2}(g) \rightarrow 2\operatorname{Li}_{3}\operatorname{N}(s) \) Now, both sides of the equation have 6 lithium (Li) atoms and 2 nitrogen (N) atoms. The final balanced equation for reaction (a) is: \(6\operatorname{Li}(s)+\mathrm{N}_{2}(g) \rightarrow 2\operatorname{Li}_{3}\operatorname{N}(s) \)
04

(b) Identify the product and its stoichiometry

For the reaction between rubidium (Rb) and sulfur (S), the product will be a rubidium sulfide compound. Since rubidium has a +1 charge and sulfur has a -2 charge, the compound formed will be Rb2S, with a balanced ratio of charges.
05

(b) Write the unbalanced equation

Now, we can write the unbalanced equation for the reaction: \( \operatorname{Rb}(s)+\mathrm{S}(s) \rightarrow \operatorname{Rb}_{2}\operatorname{S}(s) \)
06

(b) Balance the equation

To balance the equation, we need to make sure there are equal numbers of each element on both sides of the reaction. We can use the method of coefficients to achieve this: \(2 \operatorname{Rb}(s)+\mathrm{S}(s) \rightarrow \operatorname{Rb}_{2}\operatorname{S}(s) \) Now, both sides of the equation have 2 rubidium (Rb) atoms and 1 sulfur (S) atom. The final balanced equation for reaction (b) is: \(2\operatorname{Rb}(s)+\mathrm{S}(s) \rightarrow \operatorname{Rb}_{2}\operatorname{S}(s) \)

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