What volume in ml of 0.500 M silver nitrate is required to completely react with 25.0 ml of 1.5 M calcium chloride?

2AgNO3 + CaCl2 ==> 2AgCl + 2NaNO3

mols CaCl2 = M x L = ?
Using the coefficients in the balanced equation, convert mols CaCl2 to mols AgNO3. mols AgNO3 - 2x mols CaCl2.
Then M AgNO = mols AgNO3/L AgNO3. Ou know mols and M, solve for L and convert to mL.

To determine the volume of 0.500 M silver nitrate required to react with 25.0 ml of 1.5 M calcium chloride, we need to use the balanced chemical equation for the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2).

The balanced chemical equation for this reaction is:

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

From the balanced chemical equation, it can be observed that the stoichiometric ratio between silver nitrate and calcium chloride is 2:1.

Let's calculate the number of moles of calcium chloride in 25.0 ml of 1.5 M calcium chloride solution:

Number of moles of CaCl2 = volume (in liters) × molarity
= 25.0 ml × (1 L / 1000 ml) × 1.5 M
= 0.0375 mol

Since the stoichiometric ratio between silver nitrate and calcium chloride is 2:1, we need twice the number of moles of calcium chloride in silver nitrate.

Number of moles of AgNO3 = 2 × number of moles of CaCl2
= 2 × 0.0375 mol
= 0.075 mol

Now, let's calculate the volume of 0.500 M silver nitrate solution required to obtain 0.075 mol of AgNO3:

Volume (in liters) = number of moles / molarity
= 0.075 mol / 0.500 M
= 0.150 L

Finally, let's convert the volume from liters to milliliters:

Volume (in ml) = volume (in liters) × (1000 ml / 1 L)
= 0.150 L × 1000 ml
= 150 ml

Therefore, 150 ml of 0.500 M silver nitrate solution is required to completely react with 25.0 ml of 1.5 M calcium chloride.

To find the volume of silver nitrate necessary to react with calcium chloride, we can use the equation:

CaCl2 + 2AgNO3 -> 2AgCl + Ca(NO3)2

The stoichiometry of the equation tells us that we need two moles of silver nitrate for every mole of calcium chloride.

Step 1: Calculate the number of moles of calcium chloride:
First, we need to convert the given volume of calcium chloride solution to moles. The molarity (M) of the solution is 1.5 moles per liter (1.5 M), so we can use the formula:

Moles of solute (CaCl2) = Molarity (M) × Volume (L)

Since the volume is given in milliliters, we need to convert it to liters by dividing by 1000:

Moles of CaCl2 = 1.5 M × (25.0 ml / 1000) L = 0.0375 moles

Step 2: Determine the volume of silver nitrate needed:
Since the stoichiometry of the equation is 2 moles of AgNO3 for every mole of CaCl2, we can use the formula:

Volume of AgNO3 (L) = Moles of CaCl2 / Molarity (AgNO3)

We are given the molarity of the silver nitrate solution as 0.500 M:

Volume of AgNO3 = 0.0375 moles / 0.500 M = 0.075 L

Step 3: Convert the volume of silver nitrate to milliliters:
Finally, convert the volume of silver nitrate from liters to milliliters by multiplying by 1000:

Volume of AgNO3 = 0.075 L × 1000 = 75 ml

Therefore, 75 ml of 0.500 M silver nitrate is required to completely react with 25.0 ml of 1.5 M calcium chloride.