A solution containing 2.675 gram of CU CL 3 into 6 NS 3 is passed through a cation exchanger the chloride ions obtained in solution Where are treated with excess of agn O3 to give 4.78 gram of agcl the formula of the complex is

To determine the formula of the complex formed in this scenario, we need to analyze the chemical reactions and the stoichiometry involved. Let's break down the problem step by step.

Understanding the Components

We start with a solution containing 2.675 grams of copper(III) chloride, or CuCl₃. When this solution is passed through a cation exchanger, the chloride ions (Cl^-) are released into the solution. The cation exchanger typically exchanges cations for other cations in the solution, leaving behind the anions.

Formation of Silver Chloride

Next, the chloride ions in the solution are treated with an excess of silver nitrate (AgNO₃). The reaction between chloride ions and silver nitrate produces silver chloride (AgCl) as a precipitate:

• Ag⁺ + Cl^- → AgCl (s)

From the problem, we know that 4.78 grams of AgCl are formed. To find out how many moles of AgCl this represents, we can use the molar mass of AgCl.

Calculating Moles of Silver Chloride

The molar mass of AgCl is approximately 143.32 g/mol (107.87 g/mol for Ag and 35.45 g/mol for Cl). We can calculate the moles of AgCl produced:

• Moles of AgCl = Mass / Molar Mass = 4.78 g / 143.32 g/mol ≈ 0.0334 moles

Relating Chloride Ions to Copper(III) Chloride

Since each mole of AgCl corresponds to one mole of Cl^-, we can conclude that 0.0334 moles of Cl^- were produced. Now, we need to relate this back to the original copper(III) chloride.

Determining Moles of Copper(III) Chloride

The molar mass of CuCl₃ is approximately 134.45 g/mol (63.55 g/mol for Cu and 3 × 35.45 g/mol for Cl). Now, we can calculate the moles of CuCl₃ in the original solution:

• Moles of CuCl₃ = Mass / Molar Mass = 2.675 g / 134.45 g/mol ≈ 0.0199 moles

Stoichiometry of the Reaction

In copper(III) chloride, each formula unit contains three chloride ions. Therefore, the moles of chloride ions produced from the moles of CuCl₃ can be calculated as follows:

• Moles of Cl^- from CuCl₃ = 3 × Moles of CuCl₃ = 3 × 0.0199 ≈ 0.0597 moles

Finding the Complex Formula

Now we have the moles of chloride ions produced (0.0334 moles) and the moles of chloride ions that could theoretically come from the copper(III) chloride (0.0597 moles). Since we only produced 0.0334 moles of Cl^-, it indicates that not all chloride ions were released from CuCl₃. This suggests that some of the copper ions are likely forming a complex with the remaining chloride ions.

Given that copper(III) typically forms complexes, we can deduce that the complex likely involves the remaining chloride ions. The stoichiometry suggests that the complex could be represented as CuCl₂ (since two chloride ions are still associated with the copper ion). Thus, the formula of the complex formed is likely CuCl₂.

Final Thoughts

In summary, through stoichiometric calculations and understanding the reactions involved, we determined that the complex formed in this scenario is CuCl₂. This analysis highlights the importance of chemical reactions and stoichiometry in understanding complex formation in coordination chemistry.