Question

Balance the following chemical equation: $$\mathrm{CuSO}_{4}(a q)+\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s)+\mathrm{I}_{2}(s)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)$$

Short answer

The balanced chemical equation is: \[2\mathrm{CuSO}_4(aq) + 6\mathrm{KI}(s) \rightarrow 2\mathrm{CuI}(s) + 2\mathrm{I}_2(s) + 3\mathrm{K}_2\mathrm{SO}_4(aq)\].

Step-by-step solution

Write down the unbalanced equation

We are given the unbalanced chemical equation as: \[\mathrm{CuSO}_4(aq) + \mathrm{KI}(s) \rightarrow \mathrm{CuI}(s) + \mathrm{I}_2(s) + \mathrm{K}_2\mathrm{SO}_4(aq)\]

Count the number of atoms for each element on both sides of the equation

On the left side, we have: - 1 Cu atom - 1 S atom - 4 O atoms - 1 K atom - 1 I atom On the right side, we have: - 1 Cu atom - 1 S atom - 4 O atoms - 2 K atoms - 3 I atoms

Balance the Iodine (I) atoms

On the left side of the equation, we have only 1 I atom from KI. On the right side of the equation, there are 3 I atoms, in CuI and in I₂. We need to have the same number of I atoms on both sides. To do this, we can add a coefficient of 3 in front of KI, so that it becomes 3KI. This will give us 3 I atoms on the left side of the equation. The updated equation becomes: \[\mathrm{CuSO}_4(aq) + 3\mathrm{KI}(s) \rightarrow \mathrm{CuI}(s) + \mathrm{I}_2(s) + \mathrm{K}_2\mathrm{SO}_4(aq)\]

Balance the Potassium (K) atoms

Next, we need to balance the K atoms. On the left side, we now have 3 K atoms (from 3KI), and on the right side, we have 2 K atoms (from K₂SO₄). To balance this element, we can add a coefficient of 2 in front of KI and 3/2 in front of K₂SO₄. However, since using fractions is not desirable in chemical equations, we will instead multiply all the coefficients by 2 to get whole numbers. So our updated equation looks like this: \[2\mathrm{CuSO}_4(aq) + 6\mathrm{KI}(s) \rightarrow 2\mathrm{CuI}(s) + 2\mathrm{I}_2(s) + 3\mathrm{K}_2\mathrm{SO}_4(aq)\]

Balance any other elements if needed and verify that all elements are balanced

At this point, it appears our Cu, S, and O atoms are balanced. Let's verify the number of atoms for each element one more time. On the left side, there are: - 2 Cu atoms (from 2CuSO₄) - 2 S atoms (from 2CuSO₄) - 8 O atoms (from 2CuSO₄) - 6 K atoms (from 6KI) - 6 I atoms (from 6KI) On the right side, there are: - 2 Cu atoms (from 2CuI) - 2 S atoms (from 3K₂SO₄) - 8 O atoms (from 3K₂SO₄) - 6 K atoms (from 3K₂SO₄) - 6 I atoms (from 2CuI and 2I₂) Since all the elements are balanced, our balanced chemical equation is: \[2\mathrm{CuSO}_4(aq) + 6\mathrm{KI}(s) \rightarrow 2\mathrm{CuI}(s) + 2\mathrm{I}_2(s) + 3\mathrm{K}_2\mathrm{SO}_4(aq)\]

Key concepts

Stoichiometry

Stoichiometry allows us to understand the quantitative relationships in a chemical reaction. It involves using the coefficients in a balanced chemical equation to determine the amount of reactants needed or products formed. In our equation, balancing involves adjusting coefficients to ensure that each element is equally represented on both sides.

Consider the unbalanced equation: \[ \mathrm{CuSO}_4(aq) + \mathrm{KI}(s) \rightarrow \mathrm{CuI}(s) + \mathrm{I}_2(s) + \mathrm{K}_2\mathrm{SO}_4(aq) \] For stoichiometry, we start by counting the atoms and then adjust coefficients appropriately. Here, it's important to note how iodine (I) was adjusted using stoichiometric calculations to ensure both sides were balanced.

• Initially: 1 I atom (LHS), 3 I atoms (RHS)
• Solution: Add coefficient 3 to KI (LHS) to achieve 3 I atoms

Understanding stoichiometry aids in predicting reactant pathways and determining the efficiency of reactions.

Chemical Reactions

Chemical reactions describe how substances interact to form new products. This process is represented by a chemical equation, showing reactants transforming into products.

In our example, the reaction describes copper sulfate \((\mathrm{CuSO}_4)\) and potassium iodide \((\mathrm{KI})\) reacting to form copper iodide \((\mathrm{CuI})\), iodine \((\mathrm{I}_2)\), and potassium sulfate \((\mathrm{K}_2\mathrm{SO}_4)\).

• Reactants: \(\mathrm{CuSO}_4\) and \(\mathrm{KI}\)
• Products: \(\mathrm{CuI}, \mathrm{I}_2, \mathrm{K}_2\mathrm{SO}_4\)

Reactants are substances we start with, and products are the new substances formed. Properly understanding reactions is crucial in engineering, food processing, and medicine, where precise conversions can make all the difference.

Conservation of Mass

The Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction. This principle ensures that the total mass of reactants equals that of the products, guiding us to balance equations accurately.

In our example, the balancing task ensures the same number of each type of atom on both sides of the equation: - On the left side, we rearranged the original reactants to ensure that every atom initially present is accounted for in the products, thus upholding mass conservation as:

• Initial: 1 I atom (LHS), 3 I atoms (RHS)
• Matching the count involves adjusting coefficients to get equal total on both sides

By ensuring equations reflect this balance, we confirm that no atoms suddenly appear or vanish, which holds true for all chemical reactions.

Chemical Elements

Chemical Elements are the building blocks of matter, each represented by a unique symbol in the periodic table. Every chemical reaction involves these elements rearranging into different compounds.

In the reaction we're examining, various elements like Copper \((\mathrm{Cu})\), Sulfur \((\mathrm{S})\), Oxygen \((\mathrm{O})\), Potassium \((\mathrm{K})\), and Iodine \((\mathrm{I})\) are involved. These need to be balanced in terms of their atomic counts across reactants and products.- Elements on left (reactants) should match right (products)- This alignment ensures the reaction is plausible and respects chemical principlesUnderstanding these elements and their behavior during reactions helps us predict products, understand periodic trends, and manipulate compounds for various applications.