A 1.0 L sample of an aqueous solution contains .1 mol of NaCl and .1 mole of CaCl2. What is the minimum number of moles of AgNO3 that must be added to the solution in order to precipitate all of the Cl(1- charge) as AgCl?(assume that AgCl is insoluble.)
Cl^- in NaCl = 0.1 mol
Cl^- in CaCl2 = 0.1 mol x 2 = 0.2 mol
Total Cl^- = 0.3 mol/1 L = 0.3 M
AgNO3 + Cl^- ==> AgCl + NO3^-
Cl^- = 0.3 mols
So what must AgNO3 be to react with all 0.3 mol Cl^-?
Well, I'm glad you asked because I happen to be quite the chemistry comedian! To solve this problem, we need to find the limiting reagent, which is the one that will be completely consumed and determine the amount of AgNO3 needed to precipitate all the Cl(-) ions.
Now, let's see. NaCl has 1 Cl(-) ion per formula unit, so we have 0.1 moles of Cl(-) from the NaCl. CaCl2 has 2 Cl(-) ions per formula unit, so we have 0.1 moles of CaCl2 multiplied by 2, giving us a total of 0.2 moles of Cl(-) ions.
To convert moles of Cl(-) to moles of AgNO3, we need a balanced equation for the reaction. Luckily, I've got one for you:
2AgNO3 + CaCl2 => 2AgCl + Ca(NO3)2
According to the balanced equation, 2 moles of AgNO3 are needed to react with 1 mole of CaCl2, which means that 1 mole of AgNO3 is required to precipitate 1 mole of Cl(-) ions. Therefore, we need 0.2 moles of AgNO3 to react with all the Cl(-) ions.
So, the minimum number of moles of AgNO3 that must be added to the solution is 0.2. But remember, I'm here to make you smile, not just provide answers. So, let's be safe and add a few more moles just to ensure we've got all that clownishly mustache-twirling AgCl!
To determine the minimum number of moles of AgNO3 needed to precipitate all of the Cl^- ions as AgCl, we need to calculate the number of moles of Cl^- ions present in the solution.
Given that the solution contains 0.1 mol of NaCl and 0.1 mol of CaCl2, we can calculate the total number of moles of Cl^- ions in the solution.
The molar mass of NaCl is 58.44 g/mol, which means that 0.1 mol of NaCl contains:
0.1 mol NaCl x (58.44 g/mol) = 5.844 g NaCl.
The molar mass of CaCl2 is 110.98 g/mol, which means that 0.1 mol of CaCl2 contains:
0.1 mol CaCl2 x (110.98 g/mol) = 11.098 g CaCl2.
Adding the masses of NaCl and CaCl2 together gives us the total mass of Cl^- ions in the solution:
5.844 g NaCl + 11.098 g CaCl2 = 16.942 g Cl^-.
To convert the mass of Cl^- ions to moles, we divide by the molar mass of Cl^- (35.45 g/mol):
16.942 g Cl^- / 35.45 g/mol = 0.478 mol Cl^-.
Since AgCl has a 1:1 stoichiometric ratio with Cl^-, we need an equal amount of moles of AgNO3 to precipitate all of the Cl^- ions.
Therefore, the minimum number of moles of AgNO3 needed is 0.478 mol.
To find the minimum number of moles of AgNO3 required to precipitate all of the Cl(-1) as AgCl, we need to determine the number of moles of Cl(-1) ions present in the solution.
Given that the 1.0 L solution contains 0.1 mol of NaCl and 0.1 mol of CaCl2, we can calculate the total moles of Cl(-1) ions by adding the moles from NaCl and CaCl2.
Moles of Cl(-1) ions from NaCl = 0.1 mol (since NaCl dissociates into 1 Cl(-1) ion)
Moles of Cl(-1) ions from CaCl2 = 0.1 mol × 2 (since CaCl2 dissociates into 2 Cl(-1) ions)
Total moles of Cl(-1) ions = moles of Cl(-1) ions from NaCl + moles of Cl(-1) ions from CaCl2
= 0.1 mol + (0.1 mol × 2)
Therefore, the total moles of Cl(-1) ions present in the solution is 0.1 mol + 0.2 mol = 0.3 mol.
Since AgNO3 reacts with Cl(-1) ions to form AgCl, the stoichiometric ratio is 1:1. This means that 1 mol of AgNO3 will react with 1 mol of Cl(-1) ions to produce 1 mol of AgCl.
To precipitate all of the Cl(-1) ions, we need to add the same number of moles of AgNO3 as the moles of Cl(-1) ions in the solution, which is 0.3 mol.
Therefore, the minimum number of moles of AgNO3 that must be added to the solution is 0.3 mol.