Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride.After dissolving this mixture in water,0.500M silver nitrate is added dropwise until precipitate formation is complete. THe mass of the white precipitate formed is 0.641 g.
a)Calculate the mass percent of magnesium chloride in the mixture.
b)Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.
Convert the mass of the ppt (which is AgCl) to mols. # mols = g/molar mass.
Convert mols AgCl to mols MgCl2.
Convert mols MgCl2 to g MgCl2. g = mols x molar mass.
Convert g MgCl2 to percent.
%MgCl2 = [mass MgCl2/1.5 g]*100 = ??
b.
Look at the definition of molarity.
M = mols/L.
You know M and you know mols. Calculate L and convert to mL if required. Post your work if you get stuck.
i cant figure out my equation.
i have to many.this is waht i got:
MgNO3+MgCl+AgNO3->AgCl+MgCl2+2NO3-
I really don't think that's right?!
f
To calculate the mass percent of magnesium chloride in the mixture, we need to first determine the moles of magnesium chloride and magnesium nitrate present in the mixture.
Let's start by finding the moles of the white precipitate formed, which can be assumed to be the AgCl (silver chloride):
1. Determine the moles of AgCl:
- Ag has a molar mass of 107.87 g/mol and Cl has a molar mass of 35.45 g/mol.
- The molar mass of AgCl is the sum of their molar masses: 107.87 g/mol + 35.45 g/mol = 143.32 g/mol.
- The moles of AgCl can be calculated using the following formula:
Moles = Mass / Molar mass
Moles of AgCl = 0.641 g / 143.32 g/mol
Next, let's determine the moles of Cl in the AgCl precipitate:
2. Calculate the moles of Cl:
- Since one mole of AgCl contains one mole of Cl, the moles of Cl would be the same as the moles of AgCl we calculated in step 1.
Now, we can calculate the moles of magnesium chloride in the original mixture:
3. Determine the moles of MgCl2:
- In MgCl2, the magnesium (Mg) has a molar mass of 24.31 g/mol, and each chlorine (Cl) has a molar mass of 35.45 g/mol.
- The molar mass of MgCl2 is calculated by adding the molar masses: 24.31 g/mol + (2 * 35.45 g/mol) = 95.21 g/mol.
- The moles of MgCl2 can be found using the formula:
Moles = Mass / Molar mass
Moles of MgCl2 = (0.641 g / 143.32 g/mol) * (95.21 g/mol / 1 mol)
Finally, we can calculate the mass percent of magnesium chloride:
4. Calculate the mass percent of MgCl2:
- Mass percent = (Mass of MgCl2 / Total mass of the mixture) * 100
- Mass of MgCl2 = Moles of MgCl2 * Molar mass of MgCl2
- Total mass of the mixture is given as 1.50 g.
To determine the minimum volume of silver nitrate required to ensure complete formation of the precipitate, we need to calculate the moles of Cl in the precipitate and then use stoichiometry to determine the moles of AgNO3.
5. Calculate moles of Cl and moles of AgNO3:
- As calculated earlier, we have the moles of Cl from the AgCl (precipitate).
- The balanced equation for the reaction between AgNO3 (silver nitrate) and MgCl2 (magnesium chloride) is:
AgNO3 + MgCl2 → AgCl + Mg(NO3)2
- From the balanced equation, we can see that one mole of AgNO3 reacts with two moles of Cl (from MgCl2).
- Hence, Moles of AgNO3 = (Moles of Cl) / 2
Now we have all the necessary information to answer the questions:
a) Calculate the mass percent of magnesium chloride in the mixture:
- Plug in the values into the formula from step 4 above.
- Mass percent of MgCl2 = (Mass of MgCl2 / Total mass of the mixture) * 100
b) Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate:
- Calculate the moles of Cl in the AgCl (precipitate) as explained in step 2.
- Use stoichiometry to determine the moles of AgNO3 needed, as explained in step 5
- Convert moles of AgNO3 to volume using the given concentration (0.500M) and the equation:
Moles = Volume (L) * Concentration (mol/L)
Rearrange the equation to solve for Volume:
Volume = Moles / Concentration
By following these steps, you should be able to calculate the mass percent of magnesium chloride in the mixture and determine the minimum volume of silver nitrate required for complete precipitate formation.