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Step-by-Step Solution
Step 1: Calculate the mass of AgNO3 in 50 mL of its solution
The solution is 16.9% (w/v), meaning 16.9 g of AgNO3 are present in 100 mL.
Thus, in 50 mL:
$ \text{Mass of AgNO}_3 = \frac{16.9}{100} \times 50 = 8.45\text{ g} $
Step 2: Determine the number of moles of AgNO3
Molar mass of AgNO3 = $ (107.8 + 14 + 3 \times 16)\,\text{g/mol} = 169.8\,\text{g/mol} $.
Hence, number of moles:
$ n_{\text{AgNO}_3} = \frac{8.45\,\text{g}}{169.8\,\text{g/mol}} \approx 0.0497\,\text{mol} $
Step 3: Calculate the mass of NaCl in 50 mL of its solution
The solution is 5.8% (w/v), meaning 5.8 g of NaCl are present in 100 mL.
Thus, in 50 mL:
$ \text{Mass of NaCl} = \frac{5.8}{100} \times 50 = 2.9\text{ g} $
Step 4: Determine the number of moles of NaCl
Molar mass of NaCl = $ (23 + 35.5)\,\text{g/mol} = 58.5\,\text{g/mol} $.
Hence, number of moles:
$ n_{\text{NaCl}} = \frac{2.9\,\text{g}}{58.5\,\text{g/mol}} \approx 0.0495\,\text{mol} $
Step 5: Write the balanced reaction
$ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 $
From the equation, 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.
Step 6: Identify the limiting reagent
We have approximately 0.0497 mol of AgNO3 and 0.0495 mol of NaCl. These values are almost the same, with NaCl being slightly lower in moles. Thus, NaCl is the limiting reagent. This means about 0.0495 mol of AgCl will be formed.
Step 7: Calculate the mass of AgCl precipitate formed
Molar mass of AgCl = $ (107.8 + 35.5)\,\text{g/mol} = 143.3\,\text{g/mol} $.
Number of moles of AgCl formed = 0.0495 mol (based on the limiting reagent).
Hence, the mass of AgCl:
$ \text{Mass of AgCl} = n_{\text{AgCl}} \times M_{\text{AgCl}} = 0.0495 \times 143.3 \approx 7.09\,\text{g} $
Rounding suitably, the mass of the precipitate formed is about 7 g.
