Answer:
The mass of silver chloride produced = 2.202 g
Explanation:
Equation of the reaction is given below
3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)
molar mass of AgNO₃ = 170 g/mol
molar mass of FeCl₃ = 233.5 g/mol
molar mass of AgCl = 143.5 g/mol
3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate
4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃
1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃
mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1
therefore, FeCl₃ is the limiting reactant
0.0047 moles of FeCl₃ reacting will produce 0.0047 * 3 moles of AgCl = 0.0141 moles of AgCl
0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =
Therefore mass of silver chloride produced = 2.202 g
