Question

If 30.0 mL of 0.150 M CaCl₂ is added to 18.0 mL of 0.100 M AgNO_2, what is the mass of the AgCl precipitate?

Answer 1

we know that

moles = molarity x volume (ml) / 1000

so

moles of CaCl2 = 0.15 x 30 x 10-3

moles of CaCl2 = 4.5 x 10-3

now

moles of AgN02 = 0.1 x 18 x 10-3

moles of AgN02 = 1.8 x 10-3

now

the reaction is

2AgN02 + CaCl2 ---> 2AgCl + Ca(N02)2

now

we can see that

moles of CaCl2 reacted = 0.5 x moles of AgN02

so

moles of CaCl2 reacted = 0.5 x 1.8 x 10-3

moles of CaCl2 reacted = 0.9 x 10-3

but

4.5 x 10-3 moles of CaCl2 is present

so

CaCl2 is in excess and AgN02 is the limiting reagent

now

from the reaction

moles of AgCl formed = moles of AgN02 reacted = 1.8 x 10-3

now

mass = moles x molar mass

so

mass of AgCl = 1.8 x 10-3 x 143.32 = 0.258

so

0.258 grams of AgCl is precipitated out