Question

What is the mass of the precipitate formed when 50 mL of 16.9% solution of $AgNO_3$ is mixed with 50 mL of 5.8% $NaCl$ solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl =35.5)

• A. 7 g
• B. 14 g
• C. 28 g
• D. 3.5 g

Answer 1

The Correct Option is A

To solve this problem, we need to determine the mass of the precipitate formed when a solution of silver nitrate ($AgNO_3$) is mixed with a solution of sodium chloride ($NaCl$). The reaction involved is: 

\(AgNO_3 + NaCl \rightarrow AgCl \downarrow + NaNO_3\)

The precipitate formed in this reaction is silver chloride ($AgCl$).

1. Calculate the moles of $AgNO_3$ and $NaCl$.
   • The mass percentage of $AgNO_3$ solution is given as 16.9% in 50 mL of solution. This means 16.9 g of $AgNO_3$ is present in 100 mL of solution.
   • From this, calculate the amount in 50 mL: \(\text{mass of } AgNO_3 = \frac{16.9}{100} \times 50 = 8.45 \, \text{g}\).
   • The molar mass of $AgNO_3$ is 107.8 + 14 + 3×16 = 169.8 g/mol.
   • Moles of $AgNO_3 = \frac{8.45}{169.8} \approx 0.0498 \, \text{mol}$.
   • The mass percentage of $NaCl$ solution is given as 5.8% in 50 mL of solution. This means 5.8 g of $NaCl$ is present in 100 mL of solution.
   • Calculate the amount in 50 mL: \(\text{mass of } NaCl = \frac{5.8}{100} \times 50 = 2.9 \, \text{g}\).
   • The molar mass of $NaCl$ is 23 + 35.5 = 58.5 g/mol.
   • Moles of $NaCl = \frac{2.9}{58.5} \approx 0.0496 \, \text{mol}$.
2. Determine the limiting reagent.
   • The reaction is 1:1, and since moles of $AgNO_3$ (0.0498 mol) and moles of $NaCl$ (0.0496 mol) are comparable, $NaCl$ becomes the limiting reagent.
3. Calculate the amount of $AgCl$ precipitate formed.
   • The molar mass of $AgCl$ is 107.8 + 35.5 = 143.3 g/mol.
   • Mass of $AgCl$ formed = moles of $NaCl$ (since it is the limiting reagent) × molar mass of $AgCl$.
   • \(\text{Mass of } AgCl = 0.0496 \, \text{mol} \times 143.3 \, \text{g/mol} \approx 7.1 \, \text{g}\).

Therefore, the mass of the precipitate formed is approximately 7 g.

Thus, the correct option is 7 g.