CHEMISTRY PLS HELP

Question

A solution is prepared by mixing 80.0 mL of a 0.135 M NaCl solution with 24.0 mL of a 0.203 M CaCl2 solution. Determine the volume in milliliters of a 0.1313 M AgNO3 solution needed to precipitate all of the Cl− as AgCl.

volume AgNO3 =

Answer 1

The best way to start is to write all your reactants and your expected products.

AgNO3+NaCl+CaCl2 --> AgCl+Na+Ca+NO3

Note that since AgCl is the only precipitate formed, you can leave the other compounds in terms of ions in the product side (Na+,Ca2+,NO3)

The next step is to balance the reaction. I would start with balancing the Cl then go for Ag and finally NO3

3AgNO3+NaCl+CaCl2 --> 3AgCl+Na+Ca+3NO3

Now calculate the number of moles of Cl total which comes from CaCl2 and NaCl. Thus nCl=nCl from CaCl2+nCl from NaCl=2CV+CV=2(0.203M)(0.024L)+(0.135M)(0.08L)=0.021 mol. Now using the stoichometric ratio the nAgNO3=3*0.021=0.062 moles. Finally, you can calculate the volume by doing n/C=0.062mol /0.1313 M= 0.47L=470 mL