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+
Show what you can do. I'm not up to this much typing; you know this is a limiting reagent problem.
I figured out the NO3-, I'm stuck on the Cl- I have 0.24*2-0.0416= 0.4384 but I'm being told that's wrong
To determine the mass of silver chloride that can be prepared, we first need to determine the limiting reagent between silver nitrate (AgNO3) and calcium chloride (CaCl2). The limiting reagent is the one that is completely consumed in the reaction, thus determining the maximum amount of product that can be formed.
Step 1: Write the balanced equation for the reaction:
AgNO3 + CaCl2 -> AgCl + Ca(NO3)2
Step 2: Calculate the moles of each reactant using the given concentrations and volumes:
Moles of AgNO3 = concentration of AgNO3 x volume of AgNO3
Moles of AgNO3 = 0.26 M x 0.160 L
Moles of AgNO3 = 0.0416 mol
Moles of CaCl2 = concentration of CaCl2 x volume of CaCl2
Moles of CaCl2 = 0.24 M x 0.150 L
Moles of CaCl2 = 0.036 mol
Step 3: Determine the stoichiometric ratio between AgNO3 and AgCl in the balanced equation:
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of AgCl. Therefore, the ratio of moles of AgNO3 to AgCl is 1:1.
Step 4: Determine the moles of AgCl that can be formed:
The limiting reagent is determined by comparing the moles of each reactant to the stoichiometric ratio. Since the ratio is 1:1, the limiting reagent is the one with fewer moles. In this case, CaCl2 has fewer moles (0.036 mol), so it is the limiting reagent.
Step 5: Calculate the moles of AgCl that can be formed using the moles of the limiting reagent:
Moles of AgCl = moles of limiting reagent (CaCl2)
Moles of AgCl = 0.036 mol
Step 6: Convert moles of AgCl to mass of AgCl:
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 = Moles of AgCl x Molar mass of AgCl
Mass of AgCl = 0.036 mol x 143.32 g/mol
Mass of AgCl = 5.16 g
Therefore, the mass of silver chloride that can be prepared is 5.16 grams.
Now, let's move on to calculating the concentrations of each ion remaining in solution after precipitation is complete:
After the precipitation of AgCl, Ag+ and Cl- ions are consumed, while Ca2+ and NO3- ions remain in solution.
The equation for the reaction is:
AgNO3 + CaCl2 -> AgCl + Ca(NO3)2
Concentration of Cl- remaining in solution:
Since 1 mole of CaCl2 reacts with 2 moles of Cl-, the concentration of Cl- remaining in solution will be half of the initial concentration of CaCl2.
Concentration of Cl- = 0.24 M / 2 = 0.12 M
Concentration of Ag+ remaining in solution:
Since 1 mole of AgNO3 reacts with 1 mole of AgCl, the concentration of Ag+ remaining in solution will be half of the initial concentration of AgNO3.
Concentration of Ag+ = 0.26 M / 2 = 0.13 M
Concentration of Ca2+ remaining in solution:
Since Ca(NO3)2 remains in solution and does not react, the concentration of Ca2+ will be the same as the initial concentration of CaCl2.
Concentration of Ca2+ = 0.24 M
Therefore, the concentrations of each ion remaining in solution after precipitation is complete are:
Cl- = 0.12 M
Ag+ = 0.13 M
Ca2+ = 0.24 M
To find the mass of silver chloride that can be prepared, we first need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's use the stoichiometry of the reaction to find the limiting reactant:
2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
The stoichiometric ratio between silver nitrate (AgNO3) and silver chloride (AgCl) is 2:2, and the stoichiometric ratio between calcium chloride (CaCl2) and silver chloride (AgCl) is 1:2.
Step 1: Calculate the number of moles of AgNO3 and CaCl2:
Moles of AgNO3 = Concentration of AgNO3 × Volume of AgNO3 solution
= 0.26 M × 0.1600 L
= 0.0416 mol
Moles of CaCl2 = Concentration of CaCl2 × Volume of CaCl2 solution
= 0.24 M × 0.1500 L
= 0.036 mol
Step 2: Determine the limiting reactant.
The stoichiometric ratio between AgNO3 and CaCl2 is 2:1. Therefore, we need half the number of moles of AgNO3 compared to CaCl2 to react completely. Since the ratio of moles of AgNO3 (0.0416 mol) to CaCl2 (0.036 mol) is greater than 2:1, CaCl2 is the limiting reactant.
Step 3: Calculate the moles of AgCl formed:
Moles of AgCl = Moles of limiting reactant × Stoichiometric ratio
= 0.036 mol × (2 mol AgCl / 1 mol CaCl2)
= 0.072 mol
Step 4: Calculate the mass of AgCl formed:
Mass of AgCl = Moles of AgCl × Molar mass of AgCl
= 0.072 mol × 143.32 g/mol
= 10.303 g
Therefore, the mass of silver chloride that can be prepared is 10.303 grams.
Now, let's calculate the concentrations of each ion remaining in solution after the precipitation of silver chloride is complete:
The balanced equation for the reaction is:
2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
- For chloride ions (Cl‾):
Since calcium chloride is the limiting reactant, all the chloride ions will be consumed and precipitated as silver chloride. Thus, the concentration of chloride ions in the remaining solution will be zero.
- For nitrate ions (NO3‾):
For every 2 moles of silver nitrate, 2 moles of silver chloride are formed. Therefore, after the reaction, the concentration of nitrate ions in the remaining solution will remain the same as the initial concentration of silver nitrate, which is 0.26 M.
- For calcium ions (Ca2+):
Since calcium chloride is the limiting reactant, all the calcium ions will react and form calcium nitrate. Therefore, there will be no calcium ions remaining in the solution after the reaction is complete.
To summarize:
Cl‾ concentration = 0 M (all precipitated as AgCl)
NO3‾ concentration = 0.26 M
Ca2+ concentration = 0 M (all reacted to form Ca(NO3)2)