What mass of precipitate formed when 30cc of 17% solution of AgNo3 is mixed with 30cc of 5.85% NaCl solution

1)7.9g

2)10.5g

3)23g

4)4.3g

To determine the mass of precipitate formed when mixing silver nitrate (AgNO3) with sodium chloride (NaCl), we need to first understand the chemical reaction that occurs and then calculate the amounts of each reactant involved. The reaction can be represented as follows:

The Reaction

When silver nitrate reacts with sodium chloride, it forms silver chloride (AgCl), which is a precipitate, and sodium nitrate (NaNO3). The balanced chemical equation for this reaction is:

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Step 1: Calculate the moles of each reactant

First, we need to find out how many moles of AgNO3 and NaCl are present in the solutions. We can do this by using the concentration and volume of each solution.

• For AgNO3:
• Concentration = 17% w/v means 17 grams of AgNO3 in 100 mL of solution.
• Volume = 30 cc (or mL).
• Mass of AgNO3 = (17 g/100 mL) × 30 mL = 5.1 g.
• Molar mass of AgNO3 = 169.87 g/mol.
• Moles of AgNO3 = 5.1 g / 169.87 g/mol ≈ 0.0300 moles.
• For NaCl:
• Concentration = 5.85% w/v means 5.85 grams of NaCl in 100 mL of solution.
• Volume = 30 cc (or mL).
• Mass of NaCl = (5.85 g/100 mL) × 30 mL = 1.755 g.
• Molar mass of NaCl = 58.44 g/mol.
• Moles of NaCl = 1.755 g / 58.44 g/mol ≈ 0.0300 moles.

Step 2: Determine the limiting reactant

In this reaction, both AgNO3 and NaCl react in a 1:1 molar ratio. Since we have approximately 0.0300 moles of each reactant, neither is in excess, and both will completely react with each other.

Step 3: Calculate the mass of precipitate formed

The precipitate formed is silver chloride (AgCl). To find the mass of AgCl produced, we first need to calculate the moles of AgCl formed, which will be equal to the moles of the limiting reactant (either AgNO3 or NaCl, since they are equal).

• Moles of AgCl = 0.0300 moles (from the reaction).
• Molar mass of AgCl = 143.32 g/mol.
• Mass of AgCl = 0.0300 moles × 143.32 g/mol ≈ 4.30 g.

Final Result

The mass of precipitate formed when mixing the two solutions is approximately 4.30 grams. Therefore, the correct answer is 4.3 g.