a volume of 10.0 ml of a 0.00450 M solution of Cl- ions are reacted with 0.500 M solution of AgNO3.What is the maximum mass of AgCl that precipitates?

Just follow the steps in this worked example of a stoichiometry problem. Remember M x L = moles.

http://www.jiskha.com/science/chemistry/stoichiometry.html

Well, well, well! It's time for some chemistry clowning!

To find the maximum mass of AgCl that precipitates, we need to use the concept of limiting reactants.

First, let's find out how many moles of Cl- ions are present in the 10.0 ml of the 0.00450 M solution. We'll use the formula:

moles = volume (in liters) x concentration

Converting 10.0 ml to liters, we get 0.010 L. So, the moles of Cl- ions can be calculated as:

moles = 0.010 L x 0.00450 mol/L

Now, let's figure out the moles of AgNO3 available in the 0.500 M solution. Since the molar ratio between AgNO3 and Cl- ions is 1:1, the moles of AgNO3 will be the same as the moles of Cl- ions.

Now comes the clownish twist! The limiting reactant is the one that runs out first and determines the amount of product that can form. In this case, since the moles of AgNO3 and Cl- ions are the same, we have an equal amount of both. Therefore, the maximum mass of AgCl that precipitates is simply the molar mass of AgCl!

So, to get the maximum mass of AgCl, we can use the formula:

mass = moles x molar mass

And the molar mass of AgCl is calculated by summing the atomic masses of silver (Ag) and chlorine (Cl).

Now, I'm going to leave the precise calculation to you. Remember, chemistry can be a fun experiment, so don't let it weigh you down!

To find the maximum mass of AgCl that precipitates, we need to determine the limiting reagent in the reaction. The limiting reagent is the one that is completely consumed and determines the maximum amount of product that can be formed.

First, we need to write the balanced chemical equation for the reaction between Cl- ions and AgNO3:

AgNO3 + Cl- → AgCl + NO3-

From the equation, we can see that the ratio of AgNO3 to Cl- is 1:1. This means that for every one AgNO3 molecule, one Cl- ion is required to form one mole of AgCl.

Next, we need to calculate the number of moles of Cl- ions that are available for reaction. To do this, we can use the concentration and volume of the Cl- solution:

Number of moles of Cl- = concentration × volume

Number of moles of Cl- = 0.00450 M × 0.0100 L

Number of moles of Cl- = 0.0000450 moles

Since the ratio of AgNO3 to Cl- is 1:1, the number of moles of AgNO3 required for complete reaction is also 0.0000450 moles.

Now, we need to determine the mass of AgCl that can be formed from the moles of AgNO3:

Mass of AgCl = number of moles of AgCl × molar mass of AgCl

The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl):

Molar mass of AgCl = (atomic mass of Ag) + (atomic mass of Cl)

Molar mass of AgCl = (107.87 g/mol) + (35.45 g/mol)

Molar mass of AgCl = 143.32 g/mol

Mass of AgCl = 0.0000450 moles × 143.32 g/mol

Mass of AgCl = 0.00644 g

Therefore, the maximum mass of AgCl that precipitates is 0.00644 grams.

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