Calculating the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when a 50 mL solution of 16.9% w/v AgNO3 is mixed with a 50 mL solution of 5.8% w/v NaCl, we need to follow several steps.

Step 1: Convert Percentages to Grams
First, we need to convert the percentages of both solutions to grams. To do this, we use the formula:

Mass of solute (in grams) = (Percentage w/v × Volume of solution (in mL))/100

For the 16.9% w/v AgNO3 solution:
Mass of AgNO3 = (16.9 × 50)/100 = 8.45 grams

For the 5.8% w/v NaCl solution:
Mass of NaCl = (5.8 × 50)/100 = 2.9 grams

Step 2: Determine the Limiting Reagent
Next, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.

The balanced chemical equation for the reaction between AgNO3 and NaCl is:
AgNO3 + NaCl → AgCl + NaNO3

From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl. The molar mass of AgNO3 is 169.87 g/mol, and the molar mass of NaCl is 58.44 g/mol. Thus, to calculate the limiting reagent, we use the formula:

Number of moles = Mass (in grams)/Molar mass

For AgNO3:
Number of moles of AgNO3 = 8.45 g/169.87 g/mol = 0.0498 mol

For NaCl:
Number of moles of NaCl = 2.9 g/58.44 g/mol = 0.0497 mol

Since the number of moles of NaCl is slightly smaller than the number of moles of AgNO3, NaCl is the limiting reagent.

Step 3: Calculate the Mass of AgCl Precipitate
Now that we have determined the limiting reagent, we can calculate the mass of AgCl precipitate formed. From the balanced chemical equation, we know that 1 mole of NaCl reacts to produce 1 mole of AgCl. Therefore, the number of moles of AgCl formed is also 0.0497 mol.

The molar mass of AgCl is 143.32 g/mol. Thus, the mass of AgCl precipitate formed is:

Mass of AgCl = Number of moles × Molar mass = 0.0497 mol × 143.32 g/mol = 7.02 grams

Therefore, the mass of the precipitate formed when the two solutions are mixed is 7.02 grams.