Question

Write a balanced equation for each of the following decomposition reactions: (a) Potassium hydrogen carbonate decomposes by heating to give solid potassium carbonate, water, and carbon dioxide gas. (b) Strontium hydrogen carbonate decomposes by heating to give solid strontium carbonate, water, and carbon dioxide gas.

Short answer

(a) $2{\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$ (b) ${\text{Sr(HCO}}_3{)}_2\to {\text{SrCO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$

Step-by-step solution

Write the Chemical Formula for Reactants

For each decomposition reaction, start by writing down the chemical formula of the reactant.(a) Potassium hydrogen carbonate is represented as ${\text{KHCO}}_3$.(b) Strontium hydrogen carbonate is represented as ${\text{Sr(HCO}}_3{)}_2$.

Determine the Products of Decomposition

Identify the products formed from the decomposition of each reactant.(a) ${\text{KHCO}}_3$ decomposes to form potassium carbonate ${\text{K}}_2{\text{CO}}_3$, water ${\text{H}}_2\text{O}$, and carbon dioxide ${\text{CO}}_2$.(b) ${\text{Sr(HCO}}_3{)}_2$ decomposes to form strontium carbonate ${\text{SrCO}}_3$, water ${\text{H}}_2\text{O}$, and carbon dioxide ${\text{CO}}_2$.

Write the Unbalanced Chemical Equation

Set up the unbalanced chemical equation for each decomposition reaction.(a) $${\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$$(b) $${\text{Sr(HCO}}_3{)}_2\to {\text{SrCO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$$

Balance the Chemical Equations

Balance the chemical equations by adjusting the coefficients.(a) To balance the elements in the equation: - There are two potassium ions on the product side, so the coefficient for ${\text{KHCO}}_3$ should be 2 to balance potassium: $$2{\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$$ - Adjust water and carbon dioxide to fit the decomposition: $$2{\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$$(b) Strontium ions and carbonate ions are already balanced, but we need to adjust water and carbon dioxide to reflect two bicarbonates decomposing: - Adjust the coefficients: $${\text{Sr(HCO}}_3{)}_2\to {\text{SrCO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$$

Confirm the Balanced Equations

Double-check the balanced equations to make sure the number of atoms for each element is the same on both sides of the reactions:(a) 2 potassium, 2 carbon, 4 oxygen from ${\text{KHCO}}_3$ are balanced by the products. The equation is correct as written:$$2{\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$$(b) 1 strontium, 2 carbon, and 6 oxygen from ${\text{Sr(HCO}}_3{)}_2$ are balanced by the products. The equation is correct as written:$${\text{Sr(HCO}}_3{)}_2\to {\text{SrCO}}_3+{\text{H}}_2\text{O}+2{\text{CO}}_2$$

Key concepts

Chemical Formula Writing

Writing chemical formulas is a fundamental skill in chemistry that helps us represent substances in a standardized format using symbols. Every chemical compound has a unique formula that indicates its compositional elements and the number of atoms or ions of each present in a molecule or ionic compound.

Let's take a closer look at writing chemical formulas, especially for compounds such as potassium hydrogen carbonate (${\text{KHCO}}_3$). This compound consists of:

• Potassium (K)
• Hydrogen (H)
• Carbonate (${\text{CO}}_3$).

The chemical formula combines these symbols to represent the complete makeup of the compound. In ${\text{KHCO}}_3$, one atom each of potassium, hydrogen, and the whole carbonate group combine to form potassium hydrogen carbonate.

Similarly, strontium hydrogen carbonate is represented as ${\text{Sr(HCO}}_3{)}_2$. This molecule involves:

• One strontium atom (Sr)
• Two hydrogen carbonate ions (each ${\text{HCO}}_3$ consisting of hydrogen, carbon, and oxygen lightly attached).

These formulas give us a clear way to communicate the composition of these substances in any chemical reaction, such as decomposition.

Balancing Chemical Equations

Balancing chemical equations is crucial for accurately representing the conservation of mass in a chemical reaction. This means ensuring that the number of atoms of each element is the same on both the reactant and product sides of the equation.

When we initially write the unbalanced equations, like:

• ${\text{KHCO}}_3\to {\text{K}}_2{\text{CO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$
• ${\text{Sr(HCO}}_3{)}_2\to {\text{SrCO}}_3+{\text{H}}_2\text{O}+{\text{CO}}_2$

They may not represent an equal number of each type of atom. The task is to adjust the coefficients, the numbers in front of molecules, to make them equal.

For instance, in the case of $2{\text{KHCO}}_3$, we require two ${\text{KHCO}}_3$ molecules to ensure the potassium ions balance with the single ${\text{K}}_2{\text{CO}}_3$ on the product side.

Similarly, strontium hydrogen carbonate is set with its existing composition:

• The strontium atoms on both sides balance each other.
• We address hydrogen and oxygen atoms by adding suitable coefficients where needed, reflecting the two molecules of hydrogen carbonate decomposing.

The coefficients help achieve balance, resulting in precise chemical representation.

Chemical Equation Analysis

Analyzing chemical equations enables us to deeply understand the transformations happening in a chemical reaction, specifically in decomposition reactions like those given in the original exercise.

These analyses require us to break down the reactions into:

• The reactants and their compositions
• The decomposition products
• Ensuring all elements and atoms are accurately balanced

For the example of potassium hydrogen carbonate:

• Reactant: ${\text{KHCO}}_3$
• Products: potassium carbonate ${\text{K}}_2{\text{CO}}_3$, water ${\text{H}}_2\text{O}$, and carbon dioxide ${\text{CO}}_2$

Each product has roles:

• Potassium carbonate ensures the potassium atoms are conserved.
• Water and carbon dioxide account for the components decomposed from the hydrogen carbonate group.

On the other hand, ${\text{Sr(HCO}}_3{)}_2$ breaks into strontium carbonate, water, and carbon dioxide. Each analysis helps validate the correctness of the balanced equation correspondingly, confirming that no atoms are lost or created, aligning with the law of conservation of mass.