To determine the weight of AgCl that will be precipitated when a solution containing 4.77 g of NaCl is mixed with a solution of 5.77 g of AgNO₃, we employ stoichiometry. The reaction between NaCl and AgNO₃ is as follows:

$ \text{NaCl}_{(aq)} + \text{AgNO}_{3_{(aq)}} \rightarrow \text{AgCl}_{(s)} + \text{NaNO}_{3_{(aq)}} $

From the equation, you can see the reaction proceeds on a 1:1 molar basis for both NaCl and AgNO₃ to AgCl.

To start, we determine the moles of NaCl and AgNO₃ using their respective molar masses (NaCl = 58.44 g/mol, AgNO₃ = 169.87 g/mol):

$ \text{Moles of NaCl} = \frac{4.77}{58.44} \approx 0.0816 \text{ mol} $

$ \text{Moles of AgNO}_{3} = \frac{5.77}{169.87} \approx 0.0340 \text{ mol} $

Since NaCl is in excess (0.0816 mol > 0.0340 mol of AgNO₃), the limiting reagent is AgNO₃. So, the amount of AgCl formed is directly proportional to the amount of AgNO₃ present.

Now, we calculate the mass of AgCl formed. AgCl has a molar mass of approximately 143.32 g/mol (Ag = 107.87 g/mol + Cl = 35.45 g/mol). Given that 0.0340 mol of AgNO₃ will react to form the same amount of AgCl (because of the 1:1 mole ratio), the mass of AgCl produced is:

$ \text{Mass of AgCl} = 0.0340 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 4.873 \, \text{g} $

Therefore, approximately 4.873 g of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of AgNO₃.