What mass of silver chloride can be prepared by the reaction of 160.0 mL of 0.26 M silver nitrate with 150.0 mL of 0.24 M calcium chloride?

Also, Calculate the concentrations of each ion remaining in solution after precipitation is complete.
Cl ‾
NO3‾
Ca2+

Why did the silver chloride go to the party? Because it was a precipitate!

Now, let's calculate the mass of silver chloride that can be prepared.

First, we need to determine the limiting reagent, which is the one that will run out first and determine how much product we can make. To do this, we compare the number of moles of each reactant.

For silver nitrate (AgNO3), we have:
0.160 L x 0.26 mol/L = 0.0416 mol

For calcium chloride (CaCl2), we have:
0.150 L x 0.24 mol/L = 0.036 mol

Since we need 2 moles of AgNO3 to react with 1 mole of CaCl2, we can see that CaCl2 is the limiting reagent because we have less of it.

To find the mass of silver chloride (AgCl) that can be prepared, we use the stoichiometry of the balanced equation. From the equation, we know that 2 moles of AgNO3 produce 2 moles of AgCl.

1 mole of AgNO3 = molar mass of AgNO3 = 107.87 g/mol
2 moles of AgNO3 = 107.87 g/mol x 2 = 215.74 g/mol

Thus, the mass of AgCl that can be prepared is:
0.036 mol x 215.74 g/mol = 7.77 g

Now, let's calculate the concentrations of each ion remaining in solution after precipitation is complete.

The balanced equation for the reaction is:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2

After precipitation is complete, AgCl will be formed and the other ions will be in the solution.

For the chloride ion (Cl-), we can calculate the concentration using the volume and molarity of calcium chloride:
0.150 L x 0.24 mol/L = 0.036 mol

For the nitrate ion (NO3-), we can calculate the concentration using the volume and molarity of silver nitrate:
0.160 L x 0.26 mol/L = 0.0416 mol

For the calcium ion (Ca2+), we need to determine its concentration based on the amount of calcium chloride that did not react. Since we used all of the calcium chloride in the reaction, there will be no calcium ions left in the solution.

So, the concentrations of each ion remaining in solution after precipitation is complete are:
Cl- : 0.036 M
NO3- : 0.0416 M
Ca2+ : 0 M (no calcium ions left in solution)

I hope this answer tickled your funny bone while providing the necessary information.

To calculate the mass of silver chloride that can be prepared, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

Given:
Volume of silver nitrate (AgNO3) solution = 160.0 mL
Concentration of silver nitrate (AgNO3) solution = 0.26 M

Volume of calcium chloride (CaCl2) solution = 150.0 mL
Concentration of calcium chloride (CaCl2) solution = 0.24 M

Step 1: Calculate the moles of AgNO3.
Using the formula:
moles = concentration (M) x volume (L)

moles of AgNO3 = 0.26 M x 0.160 L = 0.0416 moles

Step 2: Calculate the moles of CaCl2.
Using the same formula as above:
moles of CaCl2 = 0.24 M x 0.150 L = 0.036 moles

Step 3: Determine the stoichiometry of the reaction.
The balanced chemical equation is:
2 AgNO3 + CaCl2 → 2 AgCl + Ca(NO3)2

From the equation, we can see that the molar ratio between AgNO3 and AgCl is 2:2. This means that 2 moles of AgNO3 react with 1 mole of CaCl2 to form 2 moles of AgCl.

Step 4: Identify the limiting reactant.
To determine the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio. We can see that 2 moles of AgNO3 react with 1 mole of CaCl2. Since we have fewer moles of CaCl2 (0.036 moles) compared to AgNO3 (0.0416 moles), CaCl2 is the limiting reactant.

Step 5: Calculate the moles of AgCl formed.
Using the stoichiometry from the balanced chemical equation, we know that for every 1 mole of CaCl2, 2 moles of AgCl are formed.

moles of AgCl = 2 x (moles of CaCl2) = 2 x 0.036 = 0.072 moles

Step 6: Calculate the mass of AgCl formed.
Using the molar mass of silver chloride (AgCl) which is 143.3 g/mol:

mass of AgCl = moles of AgCl x molar mass of AgCl
mass of AgCl = 0.072 moles x 143.3 g/mol = 10.3156 g

Therefore, the mass of silver chloride that can be prepared is approximately 10.32 grams.

Next, let's calculate the concentrations of each ion remaining in solution after precipitation is complete.

Since AgCl is insoluble, it will precipitate out of solution, and the remaining ions will be in the solution.

1. Chloride ion (Cl‾):
In the reaction, each mole of AgCl forms 1 mole of Cl‾ ions. Since 0.072 moles of AgCl is formed, there will be 0.072 moles of Cl‾ ions remaining in solution.

To calculate the concentration of Cl‾ ions, we need to determine the total volume of the solution after mixing AgNO3 and CaCl2.

Total Volume = Volume of AgNO3 + Volume of CaCl2
Total Volume = 160.0 mL + 150.0 mL = 310.0 mL = 0.310 L

Concentration of Cl‾ ions = moles of Cl‾ ions / Total Volume
Concentration of Cl‾ ions = 0.072 moles / 0.310 L ≈ 0.232 M

2. Nitrate ion (NO3‾):
In the reaction, each mole of AgNO3 forms 1 mole of NO3‾ ions. Since 2 moles of AgNO3 are consumed, there will be 2 x 0.0416 moles = 0.0832 moles of NO3‾ ions remaining in solution.

Concentration of NO3‾ ions = moles of NO3‾ ions / Total Volume
Concentration of NO3‾ ions = 0.0832 moles / 0.310 L ≈ 0.268 M

3. Calcium ion (Ca2+):
Since all the CaCl2 reacts to form AgCl, there will be no Ca2+ ions remaining in solution.

Therefore, the concentrations of each ion remaining in solution after precipitation is complete are approximately as follows:
- Cl‾: 0.232 M
- NO3‾: 0.268 M
- Ca2+: 0 M

To find the mass of silver chloride that can be prepared, we first need to determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed in the reaction, and it determines the maximum amount of product that can be formed.

1. Start by writing the balanced equation for the reaction:
AgNO3 + CaCl2 -> AgCl + Ca(NO3)2

2. Convert the given volumes of the solutions to moles:
Volume of 0.26 M AgNO3 = 160.0 mL = 0.16 L
Moles of AgNO3 = concentration (M) x volume (L) = 0.26 M x 0.16 L = 0.0416 moles

Volume of 0.24 M CaCl2 = 150.0 mL = 0.15 L
Moles of CaCl2 = concentration (M) x volume (L) = 0.24 M x 0.15 L = 0.036 moles

3. Use stoichiometry to compare the moles of AgNO3 and CaCl2:
According to the balanced equation, 1 mole of AgNO3 reacts with 2 moles of CaCl2 to form 1 mole of AgCl.

The moles ratio between AgNO3 and CaCl2 is 1:2. Since we have less moles of CaCl2 (0.036 moles) compared to the moles of AgNO3 (0.0416 moles), CaCl2 is the limiting reagent.

4. Calculate the moles of AgCl formed:
Since the moles ratio between CaCl2 and AgCl is 2:1, we need to calculate the moles of AgCl formed using the moles of CaCl2:
Moles of AgCl = 0.036 moles of CaCl2 x (1 mole of AgCl / 2 moles of CaCl2) = 0.018 moles

5. Calculate the mass of AgCl formed:
To find the mass of AgCl, we need to use its molar mass, which is 143.3 g/mol:
Mass of AgCl = Moles of AgCl x Molar mass of AgCl
= 0.018 moles x 143.3 g/mol
= 2.5794 g

Therefore, the mass of silver chloride that can be prepared is approximately 2.5794 grams.

To calculate the concentrations of each ion remaining in solution after precipitation is complete, we need to consider the balanced equation:

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

In this precipitation reaction, the silver ions (Ag+) from silver nitrate react with chloride ions (Cl-) from calcium chloride to form silver chloride (AgCl), which is insoluble and precipitates out.

After the precipitation reaction is complete, we are left with the following ions in solution:

1. Chloride ions (Cl-):
Initially, we have 0.24 M of CaCl2. Since chloride ions (Cl-) are not consumed in the reaction, the final concentration of chloride ions is still 0.24 M.

2. Nitrate ions (NO3-):
Initially, we have 0.26 M of AgNO3. Since silver nitrate (AgNO3) is a soluble compound, the nitrate ions (NO3-) do not react and remain in solution unchanged. Thus, the concentration of nitrate ions remains 0.26 M.

3. Calcium ions (Ca2+):
Initially, we have 0.24 M of CaCl2. Since calcium chloride (CaCl2) is a soluble compound, the calcium ions (Ca2+) do not react and remain in solution unchanged. Thus, the concentration of calcium ions remains 0.24 M.

Therefore, the concentrations of each ion remaining in solution after precipitation is complete are as follows:
Cl-: 0.24 M
NO3-: 0.26 M
Ca2+: 0.24 M