Question

One mole of acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) on reaction with excess KI will liberate \(\ldots \ldots\) moles \((\mathrm{s})\) of \(\mathrm{I}_{2}\). (a) 2 (b) 3 (c) 6 (d) 7

Short answer

The correct answer is (b) 3.

Step-by-step solution

Write the Chemical Equation

When potassium dichromate (\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)) reacts with excess potassium iodide (KI), iodine (\(\mathrm{I}_2\)) is liberated. The chemical reaction can be represented as: \[\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 6\mathrm{KI} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{K}^{+} + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 3\mathrm{I}_{2}\]

Analyze the Stoichiometry

In the balanced chemical equation, one mole of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) produces 3 moles of \(\mathrm{I}_2\) when reacted with excess KI. This is determined from the coefficients in front of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and \(\mathrm{I}_2\).

Select the Correct Answer

Based on the stoichiometry of the reaction, one mole of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) results in 3 moles of \(\mathrm{I}_2\). Therefore, option (b) 3 is correct.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Thus, the mass of reactants must equal the mass of products.
To solve stoichiometry problems, one must look at balanced chemical equations to determine the mole ratios between the reactants and products. These ratios are critical because they allow chemists to predict the quantities of substances consumed and produced in a given reaction.

• First, write the balanced chemical equation for the reaction. This involves ensuring the number of atoms for each element is the same on both sides of the equation.
• Second, use the coefficients of the balanced equation to set up conversion factors. These factors help convert quantities of one substance to another based on their mole ratios.

In the original exercise, the stoichiometry demonstrates that one mole of potassium dichromate reacts to produce three moles of iodine, based on the balanced equation. Understanding these relationships is crucial for solving stoichiometry problems effectively.

Redox Reactions

Redox reactions, also known as oxidation-reduction reactions, involve the transfer of electrons between substances. These reactions are characterized by changes in the oxidation states of the reactants. In a redox reaction, one species undergoes oxidation (loses electrons) while another undergoes reduction (gains electrons).
The original exercise involves the redox reaction between potassium dichromate and potassium iodide, where iodine is liberated. In this reaction:

• Potassium dichromate acts as the oxidizing agent and gains electrons, resulting in the formation of \(\mathrm{Cr}^{3+}\) ions.
• Potassium iodide acts as the reducing agent and loses electrons, leading to the liberation of \(\mathrm{I}_2\).

Redox reactions are crucial in many chemical processes and often involve complex balancing. The changes in oxidation states help chemists determine which species are oxidized and which are reduced, thus providing insight into the reaction mechanism.

Balanced Chemical Equations

Balanced chemical equations are essential for accurately depicting chemical reactions. They ensure the conservation of mass by having the same number of each type of atom on both sides of the equation. Balancing chemical equations guarantees that the stoichiometric calculations are precise.
To balance an equation:

• List all reactants and products.
• Count the number of atoms of each element in both reactants and products.
• Adjust coefficients to ensure the number of atoms on each side is equal.

In the given chemical equation, \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 6\mathrm{KI} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{K}^{+} + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + 3\mathrm{I}_{2}\), the coefficients reflect the stoichiometry of the reaction, demonstrating that one mole of potassium dichromate reacts with six moles of potassium iodide, effectively balancing all elements involved. This ensures accurate calculations and understanding of the chemical processes taking place.