Answer to Question #70746 in General Chemistry for Emmanuel

Question

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 Msilver nitrate with 100.0 mL of 0.15 Mcalcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

Answer 1

Find how many moles of silver nitrate are in solution:
n (AgNO3) = 0.20M*0.1L=0.02 moles
Find how many moles of calcium chloride are in solution:
n (CaCl2) = 0.15M*0.1L=0.015 moles
Reaction:
2AgNO3+CaCl2  2AgCl+Ca(NO3)2
shows the ratio moles of output reagents is n(AgNO3) : n(CaCl2) = 2:1
Therefore, calcium chloride will remain in excess after the reaction.
Find how many moles of silver chloride prepared by the reaction:
2AgNO3+CaCl2  2AgCl+Ca(NO3)2
2 2
0.02 ?
n (AgCl) = (0.02*2)/2=0.02 moles
Find mass of silver chloride:
m (AgCl)=0.02* 143,32g/mole=2.87 g
After precipitation in solution will be the following ions: Ca2+, Cl-, NO3-.
Find the volume of the formed solution:
V=100.0mL+100mL=200mL=0.2L
Find the concentrations of Ca2+:
c (Ca2+)=n (Ca2+)/V, n (Ca2+)= n (CaCl2)= 0.015 moles
c (Ca2+)=0.015/0.2=0.075M
Find the concentrations of Cl-:
c (Cl-)=n (Cl-)/V, n (Cl-)=2 * n (CaCl2) - n (AgCl)= 0.01 moles
c (Cl-)=0.01/0.2=0.05M
Find the concentrations of NO3-:
c (NO3-)=n (NO3-)/V, n (NO3-)= n (AgNO3)= 0.02 moles
c (NO3-)=0.02/0.2=0.1M

Answer
The mass of silver chloride:
m (AgCl)= 2.87 g
The concentrations of each ion:
c (Ca2+)=0.075M
c (Cl-)=0.05M
c (NO3-)=0.1M