Answer to Question #70978 in General Chemistry for Jamar Ham

Question

If 25 mL of a 0.45 M AgNO3 solution is added to 55 mL of a 1.2M NaCl solution. What mass of AgCl will be precipitated.

AgNO3 (s) + NaCl(aq) ==>> AgCl(s) + NaNO3 (aq)

Answer 1

Calculate moles of both:
n=C×V
n(AgNO_3 )=0.45 mol/L×0.025 L=0.01125 mol
n(NaCl)=1.2 mol/L×0.055 L=0.066 mol
NaCl is in excess, AgNO3 is limiting reagent. Thus,
n(AgCl)=n(AgNO_3 )=0.01125 mol
As we know:
n=m/MW
So
m=n×MW
m(AgCl)=0.01125 mol×143.23 g/mol=1.6 g