Use the References to access important valees if meeded for this question For th

Question

Use the References to access important valees if meeded for this question For the following reaction, 58.1 grams of silver nitrate are allowed to react with 21.5 grams of copper(I) chloride silver nitrate(aq) + copper(II) chloride(s)silver chloride(s)+ copper(II) nitrate(ag) What is the maximum amount of silver chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

a)

mol of CuCl2 = mass/MW = 21.5/134.45 = 0.15991

mol of AgNO3 = mass/MW = 58.1/169.87 = 0.34202

ratio is

mol of Ag =0.34202 --> 0.34202 mol of AgCl

mol of Cl- = 0.15991--> 0.15991 mol of AgCl

then, Cl- limits,

max amount of AGcl = 0.15991 mol

mass = mol*MW = 0.15991*143.32 = 22.918 g of AgCl max

b)

formula of limiting reactant, is CuCl2, or net ionic will be Cl- ions

c)

excess reagent --> 0.34202 - 0.15991 = 0.18211 mol of AgNO3

mass = mol*MW = 0.18211*169.87 = 30.935 g of AgNO3 excess

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