What mass of silver chloride in grams will precipitate when 100 mL of 0.100 M hydrochloric acid is added to 68.8 mL of 0.100 M silver nitrate solution?

Answer 1

First, find how many moles in each reagent.

As a rule, Molarity (M) = number of moles/volume of solution in liters

then, number of moles = Molarity (M) x volume of solution in litres

So, number of moles of hydrochloric acid = 0.100 x 0.100 = 0.01 moles

and number of moles of silver nitrate solution = 0.100 x 0.688 = 0.0688 moles

As the amount of silver nitrate is in excess, then hydrochloric acid is the limiting reagent, which means that the maximum amount of silver chloride can be obtained when all the 0.01 moles of hydrochloric acid react.

Now, let us get the balanced equation for the reaction:

AgNO3(aq) + HCl(aq) = AgCl(s) + HNO3(aq)

It's clear from the above equation that the ratio of HCl : AgCl is 1:1

Meaning that if you react 0.01 moles of HCl with AgNO3 you will be getting 0.01 mole of AgCl.

Now, check your university Periodic Table for proper values of RAMs of each of Ag and Cl (mine are 108 and 35.5 respectively)

Therefore, the maximum amount of AgCl that can be obtained from this reaction is 0.01 x (108 + 35.5) = 1.435g of AgCl