Question

What is the maximum mass (in grams) of silver chloride that can be precipitated by mixing 50.0 mL of 0.025 M silver nitrate solution with 100.0 mL of 0.025 M sodium chloride solution? b.) Which reagent is in excess? c.) What is the concentration of chloride remaining in solution after the reaction has gone to completion?

Answer 1

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

number of moles of AgNO3 = 50.0 ml * 0.025 M

=50.0 ml * 0.025 mole /1000 ml

=1.25*10^-3 mol

moles of NaCl = 100.0 ml * 0.025 M

=100.0 ml * 0.025 mole /1000 ml

=2.5*10^-3 mol

AgNO3 is limiting agent due to following reasons:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Now calculate the moles of AgCl with the moles of limiting agent as follows:

1.25*10^-3 mol AgNO3 * 1 mole of AgCl / 1 mol AgNO3

=1.25*10^-3 mol AgCl

mass of AgCl is as follows:

molar masso fAgCl = 143.32 g/mol

1.25*10^-3 mol AgCl*143.32 g/mol

=0.17915 g or 0.18 g

NaCl is in excess.

The concentration of chloride remaining in solution after the reaction has gone to completion = 2.5*10^-3 mol-1.25*10^-3 mol

=1.25*10^-3 mol