Question

The weight of agcl precipitated when a solution containing 8.85 g of nacl

Answer 1

NaCl + AgNO3 → AgCl + NaNO3

(4.77 g NaCl) / (58.4430 g NaCl/mol) = 0.081618 mol NaCl

(5.77 g AgNO3) / (169.87323 g AgNO3/mol) = 0.033967 mol AgNO3

0.033967 mole of AgNO3 would react completely with 0.033967 mole of NaCl, but there is more NaCl present than that, so NaCl is in excess and AgNO3 is the limiting reactant.

(0.033967 mol AgNO3) x (1 mol AgCl / 1 mol AgNO3) x (143.3212 g AgCl/mol) = 4.87 g AgCl