The balanced chemical equation for the precipitation reaction between silver nitrate AgNO3 and sodium chloride NaCl is:AgNO3 aq + NaCl aq AgCl s + NaNO3 aq To determine how many moles of silver chloride AgCl can be produced from 2.5 grams of AgNO3, first, we need to find the molar mass of AgNO3:Ag = 107.87 g/molN = 14.01 g/molO = 16.00 g/molMolar mass of AgNO3 = 107.87 + 14.01 + 3 16.00 = 169.88 g/molNow, we can find the moles of AgNO3:moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3moles of AgNO3 = 2.5 g / 169.88 g/mol = 0.0147 molSince the balanced equation shows a 1:1 ratio between AgNO3 and AgCl, the moles of AgCl produced will be equal to the moles of AgNO3:moles of AgCl = 0.0147 molTherefore, 0.0147 moles of silver chloride can be produced from 2.5 grams of AgNO3 and an excess of NaCl.