Question

# What is the maximum mass (in grams) of silver chloride that can be precipitated by mixing...

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?

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

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