Question

Complete and balance each gas-evolution equation. a. \(\operatorname{HBr}(a q)+\mathrm{NiS}(s) \longrightarrow\) b. \(\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) c. \(\mathrm{HBr}(a q)+\mathrm{Na}_{2} \mathrm{S}(a q) \longrightarrow\)

Short answer

a. \(2\text{HBr}(aq) + \text{NiS}(s) \rightarrow \text{H}_2\text{S}(g) + \text{NiBr}_2(s)\), b. \(\text{NH}_4\text{I}(aq) + \text{NaOH}(aq) \rightarrow \text{NH}_3(g) + \text{H}_2\text{O}(l) + \text{NaI}(aq)\), c. \(2\text{HBr}(aq) + \text{Na}_2\text{S}(aq) \rightarrow \text{H}_2\text{S}(g) + 2\text{NaBr}(aq)\).

Step-by-step solution

- Identify the Products of Reaction a

For the reaction between hydrobromic acid (HBr) and nickel sulfide (NiS), we predict the products based on the reactivity of the elements and the solubility rules. HBr, being an acid, will likely react with NiS to form hydrogen sulfide gas (H2S) and a nickel bromide (NiBr2).

- Balance Reaction a

To balance the equation, ensure that the same number of each type of atom appears on both sides of the equation. For reaction a, the balanced equation is: \( 2\text{HBr}(aq) + \text{NiS}(s) \rightarrow \text{H}_2\text{S}(g) + \text{NiBr}_2(s) \). We have placed a coefficient of 2 in front of HBr to balance the hydrogen and bromine atoms.

- Identify the Products of Reaction b

When ammonium iodide (NHI4) reacts with sodium hydroxide (NaOH), we can expect the formation of ammonia (NH3), water (H2O), and sodium iodide (NaI). Ammonia will be the gas evolved in this reaction.

- Balance Reaction b

Following the same principles, balance the atoms in reaction b. The balanced equation is: \(\text{NH}_4\text{I}(aq) + \text{NaOH}(aq) \rightarrow \text{NH}_3(g) + \text{H}_2\text{O}(l) + \text{NaI}(aq)\). This reaction is already balanced as written since all atoms on both sides are equal in number.

- Identify the Products of Reaction c

For the reaction between hydrobromic acid (HBr) and sodium sulfide (Na2S), the products expected are hydrogen sulfide gas (H2S) and sodium bromide (NaBr).

- Balance Reaction c

Balance this equation by ensuring the number of atoms of each element is the same on each side. The balanced equation for reaction c is: \( 2\text{HBr}(aq) + \text{Na}_2\text{S}(aq) \rightarrow \text{H}_2\text{S}(g) + 2\text{NaBr}(aq) \). We add coefficients to balance the bromine and sodium atoms.

Key concepts

Chemical Reaction Balancing

Understanding how to balance chemical equations is a foundational skill in chemistry. A balanced equation has equal numbers of each type of atom on both sides, adhering to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

Take, for example, reaction a from the exercise: \( 2\text{HBr}(aq) + \text{NiS}(s) \rightarrow \text{H}_2\text{S}(g) + \text{NiBr}_2(s) \). To achieve balance, we note that there are two bromine atoms on the left, so we need two on the right, which is why we have \( \text{NiBr}_2 \) as a product. Similarly, there are two hydrogen atoms on the left in the form of HBr, which is balanced by the two hydrogens in \( \text{H}_2\text{S} \) on the right.

It's crucial for students to practice with different types of reactions, as this will hone their ability to quickly identify which coefficients to use to balance the equations. Being methodical by balancing one element at a time can prevent mistakes and confusion.

Reactivity of Elements

The reactivity of elements plays a critical role in predicting the outcome of a chemical reaction. During a reaction, like those in the given exercise, more reactive elements tend to replace less reactive ones, leading to the formation of new compounds.

Using the reactivity series, a chart that ranks elements according to their reactivity, helps in predicting the result of displacement reactions. For instance, in reaction c: \( 2\text{HBr}(aq) + \text{Na}_2\text{S}(aq) \rightarrow \text{H}_2\text{S}(g) + 2\text{NaBr}(aq) \), the sulfur from the sodium sulfide (\( \text{Na}_2\text{S} \)) replaces the bromine in hydrobromic acid (HBr), leading to the formation of hydrogen sulfide (\( \text{H}_2\text{S} \)) and sodium bromide (\( \text{NaBr} \)).

Understanding these reactivity trends is essential for predicting not just the products that form in a reaction but also their states, whether they are aqueous, solid, gas or liquid.

Solubility Rules

Solubility rules are a set of guidelines used to predict whether an ionic compound is soluble or insoluble in water. These rules are essential when working with reaction equations, especially when the reactions involve the formation of a precipitate or the evolution of a gas.

When two aqueous solutions are mixed, the solubility rules help determine if a solid will form. For example, in the balancing of reaction b (\( \text{NH}_4\text{I}(aq) + \text{NaOH}(aq) \rightarrow \text{NH}_3(g) + \text{H}_2\text{O}(l) + \text{NaI}(aq)\)), we predict that ammonia (NH3) gas will escape and water (H2O) will remain in liquid form based on their nature, while sodium iodide (NaI) remains in solution due to the high solubility of sodium salts.

By knowing these rules, students can better understand the driving forces behind chemical reactions and correctly predict the products formed. This knowledge not only aids in balancing equations but also in laboratory practice, where understanding the solubility of compounds is crucial for carrying out reactions correctly.