Answer:
The answer to your question is 0.49 moles of AgCl
Step-by-step explanation:
Data
mass of AgNO₃ = 83 g
moles of AgCl = ?
mass of NaCl = excess
Balanced chemical reaction
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
Process
1.- Calculate the molar mass of AgNO₃
AgNO₃ = 108 + 14 + (16 x 3) = 170 g
2.- Convert the grams of AgNO₃ to moles
170 g of AgNO₃ --------------- 1 mol
83 g of AgNO₃ -------------- x
x = (83 x 1) / 170
x = 0.49 moles
3.- Calculate the moles of AgCl
1 mol of AgCl ------------ 1 mol of AgNO₃
x mol of AgCl ------------ 0.49 moles of AgNO₃
x = (0.49 x 1)/1
x = 0.49 moles of AgCl