To calculate the grams of silver chloride produced when 100 g of silver nitrate is mixed with excess sodium chloride, we need to determine the limiting reactant. The balanced chemical equation for the reaction is:

AgNO3 + NaCl → AgCl + NaNO3

First, we need to convert the given mass of silver nitrate (100 g) to moles using its molar mass (169.87 g/mol):

Moles of AgNO3 = 100 g / 169.87 g/mol = 0.583 mol

Next, we need to calculate the maximum moles of silver chloride that can be produced from 0.583 mol of silver nitrate. From the stoichiometry of the balanced equation, we know that 1 mole of AgNO3 reacts with 1 mole of AgCl. Therefore, 0.583 mol of AgNO3 can produce a maximum of 0.583 mol of AgCl.

Now, we need to check if there is enough sodium chloride to react with all of the silver nitrate. Since the problem states that there is an excess of chloride, we can assume that there is more than enough sodium chloride available.

Therefore, the limiting reactant is silver nitrate, and all of it will be consumed in the reaction. The maximum moles of silver chloride produced will be 0.583 mol.

Finally, we convert the moles of silver chloride back to grams using its molar mass (143.32 g/mol):

Grams of AgCl = 0.583 mol * 143.32 g/mol = 83.68 g

Hence, when 100 g of silver nitrate is mixed with excess sodium chloride, 83.68 g of silver chloride will be produced.