Question

What mole ratio of molecular chlorine \(\left(\mathrm{Cl}_{2}\right)\) to molecular oxygen \(\left(\mathrm{O}_{2}\right)\) would result from the breakup of the compound \(\mathrm{Cl}_{2} \mathrm{O}_{7} ?\) (1) \(1: 1\) (2) \(7: 2\) (3) \(1: 3.5\) (4) \(2: 4\)

Short answer

Option (3): 1:3.5

Step-by-step solution

Determine the Compound Composition

The compound given is \(\text{Cl}_{2} \text{O}_{7}\). This compound consists of 2 chlorine atoms and 7 oxygen atoms.

Write the Decomposition Reaction

When \(\text{Cl}_{2} \text{O}_{7}\) breaks down, it produces molecular chlorine \(\text{Cl}_{2}\) and molecular oxygen \(\text{O}_{2}\). The decomposition reaction is: \(\text{Cl}_{2} \text{O}_{7} \rightarrow \text{Cl}_{2} + \frac{7}{2} \text{O}_{2}\)

Determine the Mole Ratio

From the balanced equation, one molecule of \(\text{Cl}_{2} \text{O}_{7}\) produces one molecule of \(\text{Cl}_{2}\) and \(\frac{7}{2}\) molecules of \(\text{O}_{2}\). Therefore, the mole ratio of \(\text{Cl}_{2}\) to \(\text{O}_{2}\) is 1 : 3.5.

Choose the Correct Option

The correct mole ratio of \(\text{Cl}_{2}\) to \(\text{O}_{2}\) given in the options is \(1 : 3.5\). Therefore, option (3) is the correct answer.

Key concepts

chemical reactions

A chemical reaction describes the process where substances transform into new substances. In the given exercise, we focus on a decomposition reaction. Here, a compound breaks down into its constituent elements or simpler compounds. For our example, \(\mathrm{Cl}_2 \mathrm{O}_7\) decomposes into chlorine \(\mathrm{Cl}_2\) and oxygen \(\mathrm{O}_2\). Such reactions are essential to understanding changes in matter and predicting product and reactant quantities.

stoichiometry

Stoichiometry is the calculation method chemists use to quantify the relationships between reactants and products in a chemical reaction. By using stoichiometry, we can determine the amounts of each substance involved. In our exercise, stoichiometry helps find the mole ratio of chlorine to oxygen. This detailed calculation ensures that the elements balance and the reaction adheres to the law of conservation of mass. These ratios are vital for calculating reactant or product amounts in other reactions.

molecular composition

Understanding molecular composition involves knowing the number and types of atoms in a molecule. For instance, \(\mathrm{Cl}_2 \mathrm{O}_7\) consists of 2 chlorine atoms and 7 oxygen atoms. This knowledge is crucial for predicting decomposition products. During decomposition, \(\mathrm{Cl}_2 \mathrm{O}_7\) yields molecular chlorine \(\mathrm{Cl}_2\) and oxygen \(\mathrm{O}_2\). Breaking down the compound into simpler elements shows the direct conversion from the compound to its constituents.

balanced equations

A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. This follows the law of conservation of mass. In the given exercise, the decomposition of \(\mathrm{Cl}_2 \mathrm{O}_7\) is represented as: \[\mathrm{Cl}_2 \mathrm{O}_7 \rightarrow \mathrm{Cl}_2 + \frac{7}{2} \mathrm{O}_2\]. Each side of the equation has equivalent atoms for chlorine and oxygen, confirming it is balanced. This step is essential for accurate stoichiometric calculations and understanding reaction dynamics.