Problem 20 - Chemistry An Atoms First Approach

Question

Write balanced equations describing the reaction of Sr with each of the following: \(\mathrm{O}_{2}, \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{P}_{4}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O},\) and HCl.

Step-by-Step Solution

Answer

Here are the balanced chemical equations for the reactions of Strontium (Sr) with various substances: 1. With Oxygen (O₂): \[ 2 \mathrm{Sr} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO} \] 2. With Sulfur (S): \[ \mathrm{Sr} + \mathrm{S} \longrightarrow \mathrm{SrS} \] 3. With Chlorine (Cl₂): \[ \mathrm{Sr} + \mathrm{Cl}_{2} \longrightarrow \mathrm{SrCl}_{2} \] 4. With Phosphorus (P₄): \[ 6 \mathrm{Sr} + 2 \mathrm{P}_{4} \longrightarrow 3 \mathrm{Sr}_{3}\mathrm{P}_{2} \] 5. With Hydrogen (H₂): \[ \mathrm{Sr} + \mathrm{H}_{2} \longrightarrow \mathrm{SrH}_{2} \] 6. With Water (H₂O): \[ \mathrm{Sr} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \] 7. With Hydrochloric acid (HCl): \[ \mathrm{Sr} + 2 \mathrm{HCl} \longrightarrow \mathrm{SrCl}_{2} + \mathrm{H}_{2} \]

Step 1: Reaction with Oxygen (O₂)

When Strontium reacts with Oxygen, it forms Strontium Oxide (SrO). The balanced equation will be: \[ 2 \mathrm{Sr} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO} \]

Step 2: Reaction with Sulfur (S)

When Strontium reacts with Sulfur, it forms Strontium Sulfide (SrS). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{S} \longrightarrow \mathrm{SrS} \]

Step 3: Reaction with Chlorine (Cl₂)

When Strontium reacts with Chlorine, it forms Strontium Chloride (SrCl₂). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{Cl}_{2} \longrightarrow \mathrm{SrCl}_{2} \]

Step 4: Reaction with Phosphorus (P₄)

When Strontium reacts with Phosphorus, it forms Strontium Phosphide (Sr₃P₂). The balanced equation will be: \[ 6 \mathrm{Sr} + 2 \mathrm{P}_{4} \longrightarrow 3 \mathrm{Sr}_{3}\mathrm{P}_{2} \]

Step 5: Reaction with Hydrogen (H₂)

When Strontium reacts with Hydrogen, it forms Strontium Hydride (SrH₂). The balanced equation will be: \[ \mathrm{Sr} + \mathrm{H}_{2} \longrightarrow \mathrm{SrH}_{2} \]

Step 6: Reaction with Water (H₂O)

When Strontium reacts with Water, it forms Strontium Hydroxide (Sr(OH)₂) and Hydrogen gas. The balanced equation will be: \[ \mathrm{Sr} + 2 \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \]

Step 7: Reaction with Hydrochloric acid (HCl)

When Strontium reacts with Hydrochloric acid, it forms Strontium Chloride (SrCl₂) and Hydrogen gas. The balanced equation will be: \[ \mathrm{Sr} + 2 \mathrm{HCl} \longrightarrow \mathrm{SrCl}_{2} + \mathrm{H}_{2} \]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both the reactant and product sides of a chemical reaction. This is essential due to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in an ordinary chemical reaction. To balance a chemical equation, follow these steps:

• Write the unbalanced equation by showing the reactants on the left and the products on the right.
• Count the number of atoms for each element on both sides of the equation.
• Add coefficients (whole numbers in front of compounds) to make the number of atoms of each element equal on both sides.
• Check your work to ensure that all elements are balanced and that the coefficients are in the simplest whole-number ratio.

For example, when balancing the reaction of strontium with oxygen, the equation begins as \( \mathrm{Sr} + \mathrm{O}_2 \rightarrow \mathrm{SrO} \). You balance it by using the coefficient 2 in front of \( \mathrm{Sr} \) and \( \mathrm{SrO} \), resulting in \( 2 \mathrm{Sr} + \mathrm{O}_2 \rightarrow 2 \mathrm{SrO} \). This approach ensures that both sides of the equation have two strontium atoms and two oxygen atoms.

Strontium Reactions

Strontium is an alkaline earth metal with unique chemical properties. It readily reacts with several elements to form various compounds:

• **Strontium and Oxygen:** The reaction forms strontium oxide, \( \mathrm{SrO} \). The balanced equation is \( 2 \mathrm{Sr} + \mathrm{O}_2 \rightarrow 2 \mathrm{SrO} \).
• **Strontium and Sulfur:** This reaction yields strontium sulfide, \( \mathrm{SrS} \). The balanced equation is \( \mathrm{Sr} + \mathrm{S} \rightarrow \mathrm{SrS} \).
• **Strontium and Chlorine:** Produces strontium chloride, \( \mathrm{SrCl}_2 \). The balanced equation is \( \mathrm{Sr} + \mathrm{Cl}_2 \rightarrow \mathrm{SrCl}_2 \).
• **Strontium and Water:** Strontium hydroxide and hydrogen gas are produced, with the equation \( \mathrm{Sr} + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Sr}(OH)_2 + \mathrm{H}_2 \).

Each reaction describes how strontium interacts with different non-metals, forming stable ionic compounds in typical inorganic chemical reactions.

Inorganic Chemistry

Inorganic chemistry deals with compounds that are not based on carbon-hydrogen bonds, which differentiates it from organic chemistry. It includes the study of metals like strontium and their reactions to form various useful compounds. Within inorganic chemistry, reactions and compound formation are often categorized by type, such as:

• **Synthesis Reactions:** Two or more simple substances combine to form a more complex compound, as seen with strontium oxide formation from strontium and oxygen.
• **Single Replacement Reactions:** One element replaces another in a compound, exemplified by strontium reacting with hydrochloric acid, \( \mathrm{HCl} \), to produce strontium chloride, \( \mathrm{SrCl}_2 \), and hydrogen gas.
• **Decomposition Reactions:** A compound breaks down into simpler substances, although less common with strontium's typical reactions.

Inorganic chemistry is essential for various industries, including materials science and chemical engineering. It provides foundational knowledge on how elements like strontium behave and react, guiding practical applications.