You have four beakers labeled A, B, C, and D. In beaker A, you place 100 grams of silver nitrate and enough water to make 50 mL of solution. In beaker D, you place 100 grams of KCl and enough water to make 100 mL of solution. You also pour 100 mL of water into beaker B and 200 mL of water into beaker C. Next you pour 25 mL of the solution from beaker A into beaker B, and 25 mL of the solution from beaker D into beaker C. You then mix all of the contents from beakers B and C together into a new clean beaker.

Question

You have four beakers labeled A, B, C, and D.  In beaker A, you place 100 grams of silver nitrate and enough water to make 50 mL of solution.  In beaker D, you place 100 grams of KCl and enough water to make 100 mL of solution.  You also pour 100 mL of water into beaker B and 200 mL of water into beaker C.  Next you pour 25 mL of the solution from beaker A into beaker B, and 25 mL of the solution from beaker D into beaker C.  You then mix all of the contents from beakers B and C together into a new clean beaker.

How much precipitate will form in the new beaker?

DrBob222 · Answer

A= 100 g AgNO3/50 mL

B = 100 mL H2O

C = 200 mL H2O

D = 100 g KCl/100 mL.

Now we take 25 mL (1/2 of it) A and add to B.
Then we take 25 mL (1/2 of it) D and add to C.

Then B+C are added. What do we have?
50 g AgNO3 is 1/2 of A. 50 g KCl is 1/2 of D. We had 25 mL H2O (1/2 of 50) from A + all 100 mL from B + all 200 mL from C + 50 mL (1/2 of 100) from D. All of that makes 50 g AgNO3 + 50 g KCl + 375 mL H2O if I kept things straight. You confirm this. Then do the stoichiometry. It will be a limiting reagent + you may need to look and see if the common ion makes any difference. Check my thinking. Check my arithmetic.