the chlorine in a 0.12g sample of 95% pure magnesium chloride is to be precipitated as silver chloride, calculate the volume of 0.100m of silver nitrate solution required to precipitate the chloride and give a 100% excess.

Here is a worked example of a stoichiometry problem. After you have obtained the moles of AgNO3 needed, then moles = M x L and double the volume to obtain the 100% excess.

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To calculate the volume of silver nitrate solution required to precipitate the chloride and give a 100% excess, we first need to find the number of moles of magnesium chloride in the given sample.

Step 1: Calculate the mass of magnesium chloride in the sample.
Mass of the sample = 0.12 g
Percent purity of magnesium chloride = 95%

Mass of magnesium chloride = (95/100) * 0.12 g
= 0.114 g

Step 2: Convert the mass of magnesium chloride to moles.
Molar mass of magnesium chloride (MgCl2) = atomic mass of Mg + 2 * atomic mass of Cl
= 24.31 g/mol + 2 * 35.45 g/mol
= 95.21 g/mol

Moles of magnesium chloride = Mass of magnesium chloride / Molar mass of magnesium chloride
= 0.114 g / 95.21 g/mol
= 0.001198 moles

Step 3: Determine the stoichiometry of the reaction.
The balanced equation for the reaction between magnesium chloride and silver nitrate is:

2AgNO3 + MgCl2 -> 2AgCl + Mg(NO3)2

From the balanced equation, we can see that 1 mole of magnesium chloride reacts with 2 moles of silver nitrate to produce 2 moles of silver chloride.

Step 4: Calculate the number of moles of silver nitrate required.
Moles of silver nitrate = 2 * Moles of magnesium chloride
= 2 * 0.001198 moles
= 0.002396 moles

Step 5: Convert the moles of silver nitrate to volume using the concentration.
Given concentration of silver nitrate solution = 0.100 M

Moles of silver nitrate = Volume of silver nitrate solution * Concentration of silver nitrate
Volume of silver nitrate solution = Moles of silver nitrate / Concentration of silver nitrate
= 0.002396 moles / 0.100 M
= 0.02396 L
= 23.96 mL

Therefore, the volume of 0.100 M silver nitrate solution required to precipitate the chloride and give a 100% excess is 23.96 mL.

To calculate the volume of silver nitrate solution required, we need to follow these steps:

Step 1: Calculate the moles of magnesium chloride.
Step 2: Calculate the moles of chlorine in the magnesium chloride sample.
Step 3: Calculate the moles of silver chloride formed.
Step 4: Calculate the moles of silver nitrate required.
Step 5: Convert moles of silver nitrate to volume using its molarity.

Let's go through each step in detail:

Step 1: Calculate the moles of magnesium chloride.
Given that the sample is 0.12g and has a purity of 95%, we first need to calculate the mass of pure magnesium chloride in the sample.
Mass of pure magnesium chloride = 0.12g * 0.95 = 0.114g

Next, calculate the moles of magnesium chloride using its molar mass.
Molar mass of magnesium chloride (MgCl2) = 24.31 g/mol (for Mg) + (2 * 35.45 g/mol) (for Cl) = 95.21 g/mol

Moles of magnesium chloride = Mass / Molar mass
Moles of magnesium chloride = 0.114g / 95.21 g/mol

Step 2: Calculate the moles of chlorine in the magnesium chloride sample.
Since magnesium chloride consists of two moles of chloride ions per mole of magnesium chloride, we can multiply the moles of magnesium chloride by 2 to find the moles of chlorine.
Moles of chlorine = Moles of magnesium chloride * 2

Step 3: Calculate the moles of silver chloride formed.
Since the reaction between silver nitrate and chloride ions results in the formation of one mole of silver chloride for every mole of chloride ions, the moles of silver chloride formed will be equal to the moles of chlorine.

Step 4: Calculate the moles of silver nitrate required.
To have a 100% excess of silver nitrate, we need to add an additional mole of silver nitrate for every mole of chlorine in the sample.

Moles of silver nitrate required = Moles of silver chloride formed + Moles of chlorine

Step 5: Convert moles of silver nitrate to volume using its molarity.
Given that the molarity of the silver nitrate solution is 0.100M, we can use the following equation to calculate the volume (V) of silver nitrate solution required:

Moles = Molarity * Volume (in liters)
Volume = Moles / Molarity

Now, substitute the values into the equation to calculate the volume of silver nitrate solution required.

Note: Make sure to convert moles to liters before proceeding with the calculation if necessary.

By following these steps, you can calculate the volume of silver nitrate solution required to precipitate the chloride with a 100% excess.

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