when 130.4 mL of 0.459 mol/L silver nitrate solution is mixed with 85.23mL of 0.251 mol/L aluminum chloride solution, a precipitate of silver chloride forms. Find the mass of this Precipitate.

Answer 1

3AgNO3 + AlCl3 = 3AgCl + Al(NO3)3

So, each mole of AlCl3 requires 3 moles of AgNO3.

Figure how many moles of each you have, and you can see how many moles of AgCl you will get.

The convert that to mass.

Answer 2

Note that this is a limiting reagent problem.