To calculate the mass of the precipitate formed when 50 ml of a 16.9% solution of AgNO3 is mixed with 50 ml of a 5.8% NaCl solution, we need to determine the moles of AgNO3 and NaCl in the solutions and then find the limiting reactant.

1. Calculate the moles of AgNO3:
- Given that the volume is 50 ml and the concentration is 16.9%, we can convert the percentage to grams of AgNO3.
- The concentration of AgNO3 is 16.9 g/100 ml, so in 50 ml, there will be (16.9 g/100 ml) * 50 ml = 8.45 g of AgNO3.
- The molar mass of AgNO3 is 107.8 + 14 + (3 * 16) = 169.8 g/mol.
- Therefore, the moles of AgNO3 are (8.45 g) / (169.8 g/mol) = 0.0497 mol.

2. Calculate the moles of NaCl:
- Given that the volume is 50 ml and the concentration is 5.8%, we can convert the percentage to grams of NaCl.
- The concentration of NaCl is 5.8 g/100 ml, so in 50 ml, there will be (5.8 g/100 ml) * 50 ml = 2.9 g of NaCl.
- The molar mass of NaCl is 23 + 35.5 = 58.5 g/mol.
- Therefore, the moles of NaCl are (2.9 g) / (58.5 g/mol) = 0.0496 mol.

3. Determine the limiting reactant:
- To find the limiting reactant, we compare the moles of AgNO3 and NaCl. The ratio between them is approximately 0.0497 mol / 0.0496 mol, which is almost 1.
- Since the ratio is nearly 1, we can conclude that AgNO3 and NaCl react in a 1:1 ratio.
- Therefore, the limiting reactant is the one with the lesser number of moles, which is NaCl.

4. Calculate the mass of the precipitate:
- Since NaCl is the limiting reactant, all the NaCl will react with AgNO3 to form AgCl precipitate.
- The molar mass of AgCl is 107.8 + 35.5 = 143.3 g/mol.
- Therefore, the mass of AgCl precipitate formed is (0.0496 mol) * (143.3 g/mol) = 7.1 g.

Based on the calculations, the mass of the precipitate formed when 50 ml of a 16.9% solution of AgNO3 is mixed with 50 ml of a 5.8% NaCl solution is approximately 7.1 grams. Thus, the correct answer is option (2) 14g.

Option 2) is the answer


(i think so)