Question

Balance each of the following decomposition reactions: (a) \(\mathrm{AlPO}_{4}(s) \rightarrow \mathrm{AlPO}_{3}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{SnSO}_{4}(s) \rightarrow \mathrm{SnSO}_{3}(s)+\mathrm{O}_{2}(g)\)

Short answer

(a) \( 2\mathrm{AlPO}_{4} \rightarrow 2\mathrm{AlPO}_{3} + \mathrm{O}_{2} \); (b) \( 2\mathrm{SnSO}_{4} \rightarrow 2\mathrm{SnSO}_{3} + \mathrm{O}_{2} \)."}

Step-by-step solution

Identify the Compounds and Their Formulas

For reaction (a), identify the compounds involved:- Reactant: Aluminum phosphate, \( \mathrm{AlPO}_{4} \)- Products: Aluminum phosphite, \( \mathrm{AlPO}_{3} \), and oxygen gas, \( \mathrm{O}_{2} \).For reaction (b), identify the compounds involved:- Reactant: Tin(II) sulfate, \( \mathrm{SnSO}_{4} \)- Products: Tin(II) sulfite, \( \mathrm{SnSO}_{3} \), and oxygen gas, \( \mathrm{O}_{2} \).

Write Balanced Equation for Reaction (a)

For the decomposition reaction of \( \mathrm{AlPO}_{4} \), ensure that the number of each type of atom is equal on both sides of the equation:Initial unbalanced equation:\[\mathrm{AlPO}_{4}(s) \rightarrow \mathrm{AlPO}_{3}(s) + \mathrm{O}_{2}(g)\]Check each element:- Aluminum (Al): 1 on each side- Phosphorus (P): 1 on each side- Oxygen (O): 4 on the left, distribute equally: - 3 to \( \mathrm{AlPO}_{3} \) - 2 in \( \mathrm{O}_{2} \), needs 1 molecule of \( \mathrm{O}_{2} \)Final balanced equation:\[2\mathrm{AlPO}_{4}(s) \rightarrow 2\mathrm{AlPO}_{3}(s) + \mathrm{O}_{2}(g).\]

Write Balanced Equation for Reaction (b)

For the decomposition reaction of \( \mathrm{SnSO}_{4} \), ensure to balance the number of atoms on both sides:Initial unbalanced equation:\[\mathrm{SnSO}_{4}(s) \rightarrow \mathrm{SnSO}_{3}(s) + \mathrm{O}_{2}(g)\]Check each element:- Tin (Sn): 1 on each side- Sulfur (S): 1 on each side- Oxygen (O): 4 on the left, distribute equally: - 3 to \( \mathrm{SnSO}_{3} \) - 1 to \( \mathrm{O}_{2} \)Final balanced equation:\[2\mathrm{SnSO}_{4}(s) \rightarrow 2\mathrm{SnSO}_{3}(s) + \mathrm{O}_{2}(g).\]

Verify Balancing

Verify that each type of atom is balanced between the reactants and products: For reaction (a): - Aluminum: 2 in both reactant and product - Phosphorus: 2 in both reactant and product - Oxygen: 8 in reactant, (6 + 2) in products For reaction (b): - Tin: 2 in both reactant and product - Sulfur: 2 in both reactant and product - Oxygen: 8 in reactant, (6 + 2) in products Both reactions are correctly balanced.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the mass and charge are conserved, meaning the same number and type of atoms are present on both sides of the equation. In decomposition reactions like those involving aluminum phosphate and tin(II) sulfate, compounds break down into simpler substances.

When balancing, start by identifying the compounds and their formulas. Focus on the number of each type of atom. For aluminum phosphate (\[\mathrm{AlPO}_{4}\]), we compare aluminum (Al), phosphorus (P), and oxygen (O) atoms between the reactant and product sides. This ensures conservation of mass as per the Law of Conservation of Mass.

Step-by-step, you need to:

• Identify the initial unbalanced reaction.
• Count and match the number of atoms of each element on both sides of the equation.
• Adjust coefficients in front of compounds as needed to balance the equation.

The process requires practice but becomes intuitive over time as you get more familiar with chemical symbols and their relationships.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products involved in a chemical reaction. It's like a recipe for a chemical process, ensuring that you have just the right amount of each reactant to fully react and form products.

In our decomposition reactions, stoichiometry helps in understanding the proportion needed to balance the equation correctly. For instance, in the decomposition of aluminum phosphate (\[\mathrm{AlPO}_{4}\]), we need to ensure that the ratios are correct as we go from reactants to products. This involves recognizing the relationship between the moles of each reactant and moles of products.

Key points for stoichiometry:

• Use balanced equations to determine mole ratios.
• Calculate the mass of reactants required or products formed using these ratios.
• Convert between moles and grams if necessary, as chemists often measure by mass.

Understanding stoichiometry is crucial for executing chemical reactions safely and successfully, ensuring predictability in product yields.

Aluminum Phosphate Decomposition

In the decomposition of aluminum phosphate, \[2\mathrm{AlPO}_{4}(s) \rightarrow 2\mathrm{AlPO}_{3}(s) + \mathrm{O}_{2}(g)\], the compound breaks down into aluminum phosphite and oxygen gas. Here, decomposition is a type of chemical reaction where a single compound separates into two or more simpler substances.

For aluminum phosphate, we see:

• The initial compound, \[\mathrm{AlPO}_{4}\], is solid.
• It decomposes to \[\mathrm{AlPO}_{3}\], another solid, releasing \[\mathrm{O}_{2}\] gas.
• This requires balancing to ensure oxygen atoms are accounted for in both products.

This example underlines the importance of recognizing the phase of substances involved (solid, liquid, gas) and ensuring the atomic count is precise for maintaining balance in chemical processes.

Tin(II) Sulfate Decomposition

The decomposition of tin(II) sulfate (\[2\mathrm{SnSO}_{4}(s) \rightarrow 2\mathrm{SnSO}_{3}(s) + \mathrm{O}_{2}(g)\]) involves the breakdown of the compound into tin(II) sulfite and oxygen gas. Like aluminum phosphate decomposition, this reaction showcases the transformation from one compound to multiple products.

During this process:

• The tin(II) sulfate \[\mathrm{SnSO}_{4}\] is initially a solid.
• It breaks down to form tin(II) sulfite \[\mathrm{SnSO}_{3}\] and oxygen gas \[\mathrm{O}_{2}\].
• Balancing involves ensuring tin and sulfur atoms are consistent, adjusting so oxygen from \[\mathrm{O}_{2}\] is balanced.

This example highlights conservation of atoms and energy, as no atoms are lost, only rearranged, illustrating the nature of chemical reactions involving decomposition.