Question

If excess of AgNO\(_3\) is added to 0.05M, 100ml of Tetraaquadichlorochromium(III) chloride complex aqueous solution, then the number of moles of AgCl precipitated is \( x \times 10^{-3} \) moles. Find ‘x’.

Hint:

For precipitation reactions, use stoichiometry to relate the moles of reactants to the moles of the precipitate formed.

Answer 1

Correct Answer: 15

Step 1: Write the reaction for the precipitation of AgCl.
The chloride ions (Cl\(^-\)) from the Tetraaquadichlorochromium(III) chloride complex react with silver ions (Ag\(^+\)) from AgNO\(_3\) to form AgCl, which precipitates out. The reaction is: \[ CrCl_3 \cdot 6H_2O + 3AgNO_3 \rightarrow 3AgCl + Cr(NO_3)_3 \cdot 6H_2O \]
Step 2: Calculate the moles of CrCl\(_3\).
Given that the concentration of CrCl\(_3\) is 0.05M and the volume is 100ml or 0.1L, the moles of CrCl\(_3\) are: \[ \text{moles of CrCl}_3 = 0.05 \, \text{mol/L} \times 0.1 \, \text{L} = 0.005 \, \text{mol} \]
Step 3: Use the stoichiometry of the reaction.
From the balanced reaction, 1 mole of CrCl\(_3\) gives 3 moles of AgCl. Therefore, the moles of AgCl formed are: \[ \text{moles of AgCl} = 3 \times 0.005 = 0.015 \, \text{mol} \]
Step 4: Convert the moles of AgCl to the required units.
Since the moles of AgCl are \( 0.015 \, \text{mol} \), we express it as \( 15 \times 10^{-3} \, \text{mol} \). Therefore, \( x = 15 \).