Question

What is the approximate mass of the precipitate formed when \(50\) mL of \(16.9\%\) solution of \(\text{AgNO}_3\) is mixed with \(50\) mL of \(7.45\%\) KCl solution? (Molar mass of \(\text{AgNO}_3 = 169\) g/mol, KCl \(= 74.5\) g/mol, AgCl \(= 143.3\) g/mol)

• A. \(3.5\) g
• B. \(7\) g
• C. \(14\) g
• D. \(28\) g

Hint:

Find the moles of each reactant from the mass percent in 50 mL, then use the limiting reagent.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Silver nitrate reacts with potassium chloride in a 1:1 ratio to give a precipitate of silver chloride.
\[ \text{AgNO}_3 + \text{KCl} \to \text{AgCl}\downarrow + \text{KNO}_3 \]

Step 2: Find the amount of each reactant.
A 16.9 % solution means 16.9 g in 100 mL. So 50 mL holds 8.45 g of \(\text{AgNO}_3\), which is \(\dfrac{8.45}{169} = 0.05\) mol.
A 7.45 % solution means 7.45 g in 100 mL. So 50 mL holds 3.725 g of KCl, which is \(\dfrac{3.725}{74.5} = 0.05\) mol.

Step 3: Limiting reagent.
Both are 0.05 mol, which is exactly in the 1:1 ratio, so neither is in excess. 0.05 mol of AgCl forms.

Step 4: Mass of precipitate.
\[ 0.05\times 143.3 = 7.165\text{ g} \approx 7\text{ g} \]

Final Answer:
The mass of AgCl formed is about 7 g, option (B). \[ \boxed{\approx 7\text{ g}} \]