How many grams of silver chloride can be prepared by the reaction of 122.2 mL of 0.22 M silver nitrate with 122.2 mL of 0.16 M calcium chloride?

Answer 1

Balance the equation. Find the moles of each reactant, compare them.

Balance:

2AgNO3+CaCl2 >> 2AgCl(s) + Ca(NO3)2
Note that you use 2 moles silver nitrate to each one mole of calcium chloride.

Moles each:
AgNO3: .1222*.22=.027moles
CaCl2: .1222*.16=.020 moles

so, you do not have twice as much of silver nitrate, so it is the limiting reactant.
You will use in the reaction .027moles silver nitrate, and .027/2 moles of CaCl2

moles of product: .027moles. Convert that to grams.