# Got lost when I could't factor in the correct soultion

Submitted by hokiegal007 on Sat, 03/01/2014 - 09:54

A student mixes 100.0 mL of 0.500 M AgNO3 with 100.0 mL of 0.500 M CaCl2. (8 points)

a.       Write the balanced molecular equation for the reaction.

b.      Write the net ionic equation for the reaction.

c.       How many grams of precipitate will form?

d.      What is the concentration of Ag+, NO3, Ca2+, and Cl in               the final solution (assume volumes are additive).

The precipitation reaction will essentially go to completion:

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

Net Ionic Equation:

Ag+ (aq)     +  Cl- (aq)    -->   AgCl (s)

The Ag+ ions will be the limiting reagent.  There are .05 moles of Ag+ present and .10 moles of Cl-.

Therefore, the Ag+ will be used up to prepare .05 moles of AgCl (s), leaving .05 moles of Cl- unreacted.

To calculate the mass of AgCl, multiply the number of moles of AgCl (.05) by the molar mass of AgCl.

In calculating the concentrations, divide the number of moles of an ion by the total volume in liters (.2 L)

So, for Ag+,    molarity =   0 mol / .2 L   (this is making the approximation that all of the Ag+ reacted)

for Cl- ,    molarity =  .05 mol / .2 L

for Ca2+,  molarity =  .05 mol / 2 L

for NO3-,  molarity =  .05 mol / 2 L

Oh my gosh Spock thank you so much I am going to work out the formulas to see if I get the correct answers as you did. Your the best and thank you once again.