Problem 37 - Introductory Chemistry Atoms First

Question

Balance this chemical equation and assign it a reaction type: \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{NaNO}_{3}(a q)\)

Step-by-Step Solution

Answer

The balanced chemical equation is \(\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{3~Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{6~NaNO}_{3}(a q)\), and the reaction type is Double Displacement.

Step 1: Count atoms on each side of the equation

First, count the number of each element on the reactant (left) side and the product (right) side. Reactants: Fe: 1 N: 3 O: 9 Na: 2 S: 1 Products: Fe: 2 S: 3 Na: 1 N: 1 O: 3

Step 2: Balance the atoms in the equation

Now, balance the atoms by adjusting the coefficients of the compounds: - To balance Fe, place a coefficient of 2 in front of Fe(NO3)3 on the reactants side: \[\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{NaNO}_{3}(a q)\] - Next, balance S, by adding a coefficient of 3 in front of Na2S on the reactants side: \[\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{3~Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{NaNO}_{3}(a q)\] - Finally, balance Na and NO3, by placing a coefficient of 6 in front of NaNO3 on the products side: \[\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{3~Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{6~NaNO}_{3}(a q)\] The balanced chemical equation is: \[\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{3~Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{6~NaNO}_{3}(a q)\]

Step 3: Determine the reaction type

In this reaction, the two reactants (Fe(NO3)3 and Na2S) exchange ions with each other, resulting in the formation of two new products (Fe2S3 and NaNO3). This type of reaction is known as a double displacement (or metathesis) reaction. So, the given balanced chemical equation and its reaction type are: \[\mathrm{2~Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{3~Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)+\mathrm{6~NaNO}_{3}(a q)\] \[\text{Reaction Type: Double Displacement}\]

Key Concepts

Double Displacement Reaction

A double displacement reaction, also known as a metathesis reaction, represents a type of chemical process where two compounds exchange ions to form two new compounds. Imagine it like a dance between two couples (the reactants) where partners swap to form new pairs (the products). In our textbook problem, iron(III) nitrate and sodium sulfide are the reactants that undergo this fascinating dance. By exchanging ions, iron(III) nitrate and sodium sulfide form iron(III) sulfide and sodium nitrate. The general form of a double displacement reaction can be represented by \(AB + CD \rightarrow AD + CB\), where A and C are cations, while B and D are anions. Balancing such equations requires careful attention to the number of atoms of each element on both sides of the reaction. This ensures the law of conservation of mass is upheld, meaning that the mass remains constant and atoms are neither created nor destroyed.

Chemical Reaction Types

Chemistry is filled with different types of reactions that can seem endless and overwhelming. However, most of these reactions can be categorized into a few basic types, which help students to understand and predict the outcomes of chemical processes. These categories include synthesis reactions, where substances combine to form a more complex product; decomposition reactions, where a compound breaks down into simpler substances; single displacement reactions, involving one element being replaced in a compound by another; and as we saw in the previous section, double displacement reactions.Understanding the type of reaction helps in predicting the products and the energy changes that occur. For example, combustion reactions release heat and light as products burn in the presence of oxygen, whereas endothermic reactions absorb heat from the surroundings. Identification of the reaction type is a fundamental step in solving chemical equations and is especially important when students encounter unfamiliar reactions.

Stoichiometry

Stoichiometry lies at the core of the chemical equations we balance. It is the section of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. The term stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), which aptly describes its focus on measuring the elements. When we balance equations, such as the double displacement reaction in our example, we are engaging in stoichiometry, ensuring that the number of atoms for each element is the same on both sides of the reaction equation. It's like a meticulous game of scales where the atoms and molecules must be perfectly balanced to reflect the mass conservation principle. This concept is essential not just for homework exercises but also for real-world applications such as calculating dosages for medications, determining the yield of a product in a chemical reaction, or analyzing the fuel consumption in a car engine.