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?

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.

To determine how much precipitate will form in the new beaker, we need to understand the reactions that occur when silver nitrate (AgNO3) and potassium chloride (KCl) solutions are mixed.

Silver nitrate (AgNO3) reacts with potassium chloride (KCl) to form silver chloride (AgCl) and potassium nitrate (KNO3).

The balanced chemical equation for this reaction is:

AgNO3 + KCl -> AgCl + KNO3

Since we mixed 25 mL of the solution from beaker A (silver nitrate) with 25 mL of the solution from beaker D (potassium chloride), we have a total volume of 50 mL of mixed solution.

Now, let's calculate the number of moles of each substance:

The molar mass of AgNO3 is 169.87 g/mol. We initially had 100 grams, so the number of moles of AgNO3 is:

moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 100 g / 169.87 g/mol
= 0.588 mol

The molar mass of KCl is 74.55 g/mol. We initially had 100 grams, so the number of moles of KCl is:

moles of KCl = mass of KCl / molar mass of KCl
= 100 g / 74.55 g/mol
= 1.341 mol

From the balanced chemical equation, we can see that 1 mole of AgNO3 reacts with 1 mole of KCl to produce 1 mole of AgCl.

So, the limiting reagent in this reaction is AgNO3 because we have fewer moles of AgNO3 than KCl.

Since 1 mole of AgNO3 forms 1 mole of AgCl, and we have 0.588 moles of AgNO3, the maximum amount of AgCl that can form is also 0.588 moles.

To calculate the mass of AgCl precipitate that forms, we multiply the number of moles by the molar mass of AgCl.

mass of AgCl = moles of AgCl x molar mass of AgCl
= 0.588 mol x 143.32 g/mol
= 84.26 g

Therefore, approximately 84.26 grams of silver chloride (AgCl) precipitate will form in the new beaker.