Question

A 450.0 mL sample of a 0.295 M solution of silver nitrate is mixed with 400.0 mL of 0.200 M calcium chloride. What is the concentration of Cl- in solution after the reaction is complete?

Answer 1

volume of AgNO3, V = 450.0 mL

= 0.45 L

we have below equation to be used:

number of mol in AgNO3,

n = Molarity * Volume

= 0.295*0.45

= 0.1328 mol

volume of CaCl2, V = 400.0 mL

= 0.4 L

we have below equation to be used:

number of mol in CaCl2,

n = Molarity * Volume

= 0.2*0.4

= 8*10^-2 mol

we have the Balanced chemical equation as:

2 AgNO3 + CaCl2 ---> 2 AgCl + Ca(NO3)2

2 mol of AgNO3 reacts with 1 mol of CaCl2

for 0.1328 mol of AgNO3, 6.638*10^-2 mol of CaCl2 is required

But we have 8*10^-2 mol of CaCl2

so, AgNO3 is limiting reagent

From balanced chemical reaction, we see that

when 2 mol of AgNO3 reacts, 1 mol of CaCl2 reacts

mol of CaCl2 reacted = (1/2)* moles of AgNO3

= (1/2)*0.1328

= 6.638*10^-2 mol

mol of CaCl2 remaining = mol initially present - mol reacted

mol of CaCl2 remaining = 8*10^-2 - 6.638*10^-2

mol of CaCl2 remaining = 1.363*10^-2 mol

total volume now = 450.0 mL + 400.0 ml = 850.0 mL

[CaCl2] remaining = mol of CaCl2 remaining / volume

= 1.363*10^-2 mol / 0.850 L

= 0.0160 M

[Cl-] remaining = 2* 0.0160 M

= 0.0320 M

Answer: 0.0320 M