Question

Complete and balance the equation for each of the following decomposition reactions. (a) \(\mathrm{CaCO}_{3}(s) \stackrel{\text { leat }}{\longrightarrow}\) (b) \(\mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\text { heat }}{\longrightarrow}\)

Short answer

The balanced decomposition equation for (a) is \(\mathrm{CaCO}_{3}(s) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(s) + \mathrm{CO}_{2}(g)\) and for (b) is \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\text { heat }}{\longrightarrow}\mathrm{CuSO}_{4}(s) + 5\mathrm{H}_{2}O(l)\)

Step-by-step solution

Identify the products for equation (a)

In the decomposition of Calcium Carbonate \(\mathrm{CaCO}_{3}(s) \), the products are Calcium Oxide (\(\mathrm{CaO}(s)\)) and Carbon Dioxide (\(\mathrm{CO}_{2}(g)\)). So we write the equation as: \(\mathrm{CaCO}_{3}(s) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(s) + \mathrm{CO}_{2}(g)\)

Balance the equation for (a)

The equation \(\mathrm{CaCO}_{3}(s) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(s)+ \mathrm{CO}_{2}(g)\) is already balanced. The Calcium, Carbon and Oxygen atoms are equal on both sides of the equation. So we can move on to the next decomposition equation.

Identify the products for equation (b)

In the decomposition of Copper(II) Sulfate pentahydrate \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s)\), the products are Copper(II) Sulfate (\(\mathrm{CuSO}_{4}(s)\)) and Water (\(\mathrm{H}_{2}O(l)\)). So we write the equation as: \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\text { heat }}{\longrightarrow}\mathrm{CuSO}_{4}(s) + \mathrm{H}_{2}O(l)\)

Balance the equation for (b)

In this equation, there are 5 water molecules result from the decomposition. So balance the water molecules on the right side of the equation. This gives us: \(\mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \stackrel{\text { heat }}{\longrightarrow}\mathrm{CuSO}_{4}(s) + 5\mathrm{H}_{2}O(l)\)

Key concepts

Chemical Equation Balancing

When working on chemical reactions, it's essential to balance the chemical equations to adhere to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms for each element must be the same on both sides of the equation.

To balance a chemical equation, one must ensure that the number of atoms of each element is equal on both reactant and product sides. In our example (a), the decomposition of Calcium Carbonate, \( \mathrm{CaCO}_{3}(s) \), forms Calcium Oxide, \( \mathrm{CaO}(s) \), and Carbon Dioxide, \( \mathrm{CO}_{2}(g) \). The equation is already balanced, indicating an equal number of calcium, carbon, and oxygen atoms before and after the reaction. In case an equation is not balanced, coefficients should be added before the chemical formulas to equalize the atom counts.

Here are steps you can follow to balance chemical equations:

• Write down the number of atoms of each element present in the unbalanced equation.
• Adjust coefficients to get the same number of atoms of each element on both sides.
• Start by balancing elements that appear in only one reactant and one product.
• Balance hydrogen and oxygen atoms last, as they often appear in multiple compounds.
• Check your work to ensure both sides of the equation are balanced.

Thermal Decomposition

One class of chemical reactions is thermal decomposition, where a single compound breaks down into two or more products when heated. This process is a type of endothermic reaction, meaning it absorbs energy in the form of heat.

In our examples, both Calcium Carbonate and Copper(II) Sulfate pentahydrate undergo thermal decomposition. Heat acts as the catalyst, causing Calcium Carbonate to form Calcium Oxide and Carbon Dioxide, while Copper(II) Sulfate pentahydrate decomposes into Copper(II) Sulfate and Water. These reactions are important in various industrial processes, including the production of lime from limestone and the refinement of minerals.

Factors Affecting Thermal Decomposition

• Temperature: Higher temperatures generally increase the rate of decomposition.
• Compound Stability: Some compounds require more energy to break their bonds due to their stability.
• Particle Size: Smaller particles may decompose quicker due to their larger surface area.
• Purity: Impurities can either inhibit or promote decomposition.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical reactions, involving the calculation of relative quantities of reactants and products in chemical reactions. Stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), suggesting the measurement of elements.

Through stoichiometry, one can predict the amount of products produced or reactants needed for a given reaction. In the decomposition reactions we discussed, stoichiometry allows us to directly relate the amount of starting material to the amounts of products formed. For instance, from the balanced equation for the decomposition of Copper(II) Sulfate pentahydrate, \( \mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \), stoichiometry tells us that 5 moles of water (\(5\mathrm{H}_{2}O\)) will result from the decomposition of 1 mole of \( \mathrm{CuSO}_{4}\cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \).

Understanding the stoichiometric coefficients in a balanced equation is crucial for determining the ratios in which substances react and form products. The principles of stoichiometry also apply to the conservation of mass and charge, as well as the energy balance in a reaction.