Problem 23 - Chemistry The Central Science In Si Units

Question

Complete and balance the following equations: (a) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\) (d) \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\) (e) \(\mathrm{PbO}(s)+\mathrm{H}_{2}(g) \longrightarrow\)

Step-by-Step Solution

Answer

(a) \(\mathrm{NaH} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH} + \mathrm{H}_{2}\); (b) \(\mathrm{Fe} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{FeSO}_{4} + \mathrm{H}_{2}\); (c) \(\mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{HBr}\); (d) \(2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\); (e) \(\mathrm{PbO} + \mathrm{H}_{2} \rightarrow \mathrm{Pb} + \mathrm{H}_{2} \mathrm{O}\).

Step 1: Identifying the Reaction Type and Products for Equation (a)

The reaction involves sodium hydride \(\mathrm{NaH}\) reacting with water \(\mathrm{H}_{2} \mathrm{O}\). This is a displacement reaction where the products are sodium hydroxide \(\mathrm{NaOH}\) and hydrogen gas \(\mathrm{H}_{2}\), because the hydride ion \((\mathrm{H}^-)\) will react with water to displace hydrogen. Thus, the reaction can be written as: \[\mathrm{NaH} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH} + \mathrm{H}_{2}.\]

Step 2: Balancing the Equation (a)

Checking the number of each type of atom on both sides of the equation, we see that there is one sodium atom, two hydrogen atoms, and one oxygen atom on both sides. The equation is already balanced as \[\mathrm{NaH} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH} + \mathrm{H}_{2}.\]

Step 3: Identifying the Reaction Type and Products for Equation (b)

Iron \(\mathrm{Fe}\) reacts with sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to displace hydrogen and form iron sulfate \((\mathrm{FeSO}_{4})\) and hydrogen gas \(\mathrm{H}_{2}\). The unbalanced reaction is \[\mathrm{Fe} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{FeSO}_{4} + \mathrm{H}_{2}.\]

Step 4: Balancing the Equation (b)

The balanced equation for iron and sulfuric acid is \[\mathrm{Fe} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{FeSO}_{4} + \mathrm{H}_{2}.\] One atom of iron, two hydrogen atoms, and one sulfate group \((\mathrm{SO}_{4})\) are present on both sides, so it is balanced.

Step 5: Identifying Reaction Type and Products for Equation (c)

The reaction between hydrogen and bromine involves synthesis. The product formed is hydrogen bromide \((\mathrm{HBr})\). The initial equation is \[\mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow \mathrm{HBr}.\]

Step 6: Balancing the Equation (c)

Hydrogen bromide \((\mathrm{HBr})\) should have two molecules to balance bromine's two atoms. The balanced equation is \[\mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{HBr}.\]

Step 7: Identifying Reaction Type and Products for Equation (d)

Sodium \((\mathrm{Na})\) reacts with hydrogen \((\mathrm{H}_{2})\) and forms sodium hydride \((\mathrm{NaH})\). The preliminary chemical equation is \[2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}.\]

Step 8: Balancing the Equation (d)

With two sodium atoms and the diatomic nature of hydrogen gas, the balanced equation is \[2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}.\]

Step 9: Identifying Reaction Type and Products for Equation (e)

Lead(II) oxide \((\mathrm{PbO})\) reacts with hydrogen gas \((\mathrm{H}_{2})\) to form elemental lead \((\mathrm{Pb})\) and water \((\mathrm{H}_{2} \mathrm{O})\). The unbalanced reaction is \[\mathrm{PbO} + \mathrm{H}_{2} \rightarrow \mathrm{Pb} + \mathrm{H}_{2} \mathrm{O}.\]

Step 10: Balancing the Equation (e)

Both sides have one lead atom, two hydrogen atoms, and one oxygen atom. The equation is balanced as \[\mathrm{PbO} + \mathrm{H}_{2} \rightarrow \mathrm{Pb} + \mathrm{H}_{2} \mathrm{O}.\]

Key Concepts

Displacement Reactions

Displacement reactions, also known as single replacement reactions, are fascinating chemical processes where one element displaces another in a compound. Typically, an element in its pure state will replace a similar element that is part of a compound. This results in the formation of a new compound, while the displaced element is left in its elemental form.
For example, consider the reaction between iron (Fe) and sulfuric acid (H\(_2\)SO\(_4\)). The iron displaces hydrogen in the compound to form iron sulfate (FeSO\(_4\)) and releases hydrogen gas (H\(_2\)). The chemical equation can be written as:

• Fe + H\(_2\)SO\(_4\) → FeSO\(_4\) + H\(_2\)

Here, Fe displaces hydrogen, showcasing a classic example of a displacement reaction. In these reactions, priorities are often determined by the reactivity of the elements involved. Irons' willingness to participate can be predicted using the reactivity series of metals.

Synthesis Reactions

Synthesis reactions are intriguing chemical processes where two or more substances combine to form a single, more complex product. This type of reaction is also known as combination reaction due to its inherent nature of bringing together distinct elements or compounds.
An excellent example of a synthesis reaction is the combination of hydrogen gas (H\(_2\)) and bromine gas (Br\(_2\)) to create hydrogen bromide (HBr). This reaction is evident in the balanced equation:

• H\(_2\) + Br\(_2\) → 2HBr

During this reaction, individual atoms of hydrogen and bromine come together to form a more complex molecule, hydrogen bromide. Synthesis reactions are fundamental in building macroscopic structures from atomic and molecular levels, playing a key role in chemical manufacturing and material science.

Reaction Types

Understanding different types of chemical reactions is crucial for students and scientists alike as it helps predict the products of a given reactant set. Chemical reactions can be classified into several types: displacement, synthesis, decomposition, combustion, and more. Here, we explore two prominently featured types in our exercises - displacement and synthesis reactions.

• Displacement Reaction: As explained, one element trades places with another element in a compound. For instance, sodium (Na) reacts with hydrogen gas (H\(_2\)) to form sodium hydride (NaH).
• Synthesis Reaction: Involves the combination of two or more simple substances to form a more complex substance, like hydrogen and bromine forming hydrogen bromide (HBr).

Recognizing these reaction types allows you to understand reaction mechanisms and predict their outcomes more effectively, facilitating a comprehensive grasp of chemistry fundamentals.

Steps in Chemical Equation Balancing

Balancing chemical equations is an essential skill in chemistry. It ensures that the same number and type of atoms are present on both sides of the equation, reflecting the law of conservation of mass.
To balance a chemical equation, follow these steps:

• Identify the reactants and products in the equation.
• List how many atoms of each element are present on both the reactant and product sides.
• Adjust coefficients (the numbers before molecules) to have the same number of each type of atom on both sides of the equation.
• Check the balance several times to confirm accuracy.

For instance, in the equation NaH + H\(_2\)O → NaOH + H\(_2\), each element is already balanced with one sodium, two hydrogens, and one oxygen atom on both sides. By practicing these steps, you can accurately balance even complex equations, strengthening your chemistry skills.