You mix 1.00 L of 2.00 M BaCl2 with 1.00 L of 2.00 M AgNO3. What compounds remain in solution, and what are their concentrations?

Both compounds are soluble, thus you have a double replacement reaction. Do stochiometry with the balanced reaction to find the concentrations.

mols BaCl2 = 1 L x 2M = 2 mols.

mols AgNO3 = 1 L x 2M = 2 mols.
......BaCl2 + 2AgNO3 ==>2AgCl + Ba(NO3)2
initial.2......2.........0.......0
change.-1.....-2.........2.......1
equil...1......0.........2.......1
AgCl is a solid. You should be able to take it from here.

To determine the compounds that remain in solution and their concentrations, we need to consider the possible chemical reaction that may occur between barium chloride (BaCl2) and silver nitrate (AgNO3).

The chemical equation for the reaction between BaCl2 and AgNO3 is:

BaCl2 + 2AgNO3 -> 2AgCl + Ba(NO3)2

This equation shows that when BaCl2 reacts with AgNO3, it forms silver chloride (AgCl) and barium nitrate (Ba(NO3)2).

Since both BaCl2 and AgNO3 have the same concentration of 2.00 M and we mix equal volumes of 1.00 L, we can use the stoichiometry of the reaction to determine the amount of each reactant that reacts.

The stoichiometry of the reaction indicates that one mole of BaCl2 reacts with two moles of AgNO3. Therefore, the amount of AgNO3 available is in excess, and all the BaCl2 will react.

When BaCl2 reacts completely with AgNO3, it forms two moles of AgCl and one mole of Ba(NO3)2. However, due to the equal initial concentrations and volumes, the concentrations of the resulting compounds will also be the same.

Therefore, the remaining compounds in solution after the reaction are silver chloride (AgCl) and barium nitrate (Ba(NO3)2), and their concentrations are both 2.00 M.