Question

Write a balanced equation for each of the following combination reactions: (a) Tin metal is heated with oxygen gas to yield tin(IV) oxide. (b) Lead metal is heated with oxygen gas to yield lead(II) oxide.

Short answer

(a) \( 2\text{Sn} + \text{O}_2 \rightarrow 2\text{SnO}_2 \); (b) \( 2\text{Pb} + \text{O}_2 \rightarrow 2\text{PbO} \).

Step-by-step solution

Identify the Reactants and Products for Reaction (a)

In reaction (a), we have tin metal and oxygen gas as reactants, and the product is tin(IV) oxide. The chemical symbols for these are \( \text{Sn} \) for tin, \( \text{O}_2 \) for oxygen gas, and \( \text{SnO}_2 \) for tin(IV) oxide.

Write the Unbalanced Equation for Reaction (a)

For the combination reaction of tin with oxygen, the equation is \( \text{Sn} + \text{O}_2 \rightarrow \text{SnO}_2 \). This equation is currently not balanced.

Balance the Equation for Reaction (a)

To balance the equation for tin(IV) oxide, ensure the number of each type of atom is equal on both sides. We need two atoms of tin and two atoms of oxygen, so the balanced equation is \( 2\text{Sn} + \text{O}_2 \rightarrow 2\text{SnO}_2 \).

Identify the Reactants and Products for Reaction (b)

In reaction (b), we have lead metal and oxygen gas as reactants, and the product is lead(II) oxide. The chemical symbols are \( \text{Pb} \) for lead, \( \text{O}_2 \) for oxygen gas, and \( \text{PbO} \) for lead(II) oxide.

Write the Unbalanced Equation for Reaction (b)

For the combination reaction of lead with oxygen, the equation is \( \text{Pb} + \text{O}_2 \rightarrow \text{PbO} \). This equation is currently unbalanced.

Balance the Equation for Reaction (b)

To balance the equation for lead(II) oxide, ensure the number of each type of atom is equal on both sides. We need two units of \( \text{PbO} \) to provide two oxygen atoms. The balanced equation is \( 2\text{Pb} + \text{O}_2 \rightarrow 2\text{PbO} \).

Key concepts

Balancing Equations

Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is upheld, meaning that matter is neither created nor destroyed in a chemical reaction. Therefore, the number of atoms for each element needs to be identical on both sides of the equation.

Here is how you can balance equations:

• Identify the reactants and the products. For example, in the equations provided, tin and lead react with oxygen to produce metal oxides.
• Write the unbalanced chemical equation using correct chemical formulas. Initially, you may write it as, for example, for tin: \( \text{Sn} + \text{O}_2 \rightarrow \text{SnO}_2 \).
• Adjust the coefficients of the equation to balance it. Count the atoms of elements on both sides of the equation to make sure they match.

For instance, in the case of the tin equation \( \text{Sn} + \text{O}_2 \rightarrow \text{SnO}_2 \), the equation is not balanced because there is one tin on the left and two oxygens. By adjusting to \( 2\text{Sn} + \text{O}_2 \rightarrow 2\text{SnO}_2 \), it balances the number of tin and oxygen atoms on both sides.

Combination Reactions

Combination reactions are a type of chemical reaction where two or more reactants form a single product. This type is uncomplicated and common in chemistry, particularly when a metal combines with oxygen.

Features of combination reactions include:

• The reactants could be elements or compounds, but they produce a single, more complex product.
• They often involve oxidation-reduction processes, where one element transfers electrons to another.

In the exercise you are working with, these reactions involve metals reacting with oxygen. This leads to the formation of their respective metal oxides. For example, tin reacts with oxygen to form tin(IV) oxide \( \text{SnO}_2 \), and lead reacts with oxygen to form lead(II) oxide \( \text{PbO} \). This straightforward transformation illustrates a classic example of a combination reaction.

Oxidation

Oxidation is a fundamental concept in chemistry often observed in combination reactions, particularly when a metal reacts with oxygen, as in our current discussion. It involves the loss of electrons by a material, with oxygen commonly being the oxidizing agent.

Here’s what you need to know about oxidation:

• Oxidation does not always involve oxygen, but your exercise focuses on metal-oxygen reactions where oxygen acts as the oxidizer.
• Metals like tin \( \text{Sn} \) and lead \( \text{Pb} \) lose electrons and form positive ions when they combine with oxygen.

For instance, when tin oxidizes, it loses electrons to oxygen, forming \( \text{Sn}^{4+} \) in \( \text{SnO}_2 \) while oxygen becomes \( \text{O}^{2-} \). This change in oxidation states reflects the transfer of electrons and demonstrates the concept of oxidation and reduction happening simultaneously.