Based on the following equation: 2AgNO3 + CaCl2 > 2AgCl + Ca(NO3) When 50 g AgNO3 are combined with 1.45M CaCl2 solution, what volume of the 1.45M CaCl2 solution would be needed to precipitate all of the silver as AgCl?
mols AgNO3 = grams/molar mass = approx 0.3 but you need a better answer than that estimate.
Convert mols AgNO3 to mols CaCl2. That means approx 0.3 x (1 mol CaCl2/2 mols AgNO3) = approx 0.15 mol CaCl2.
Then M CaCl2 = mols CaCl2/L CaCl2. You have M CaCl2 and mols CaCl2, solve for L CaCl2. Covert to mL if needed.
To solve this problem, we need to use stoichiometry and the concept of balanced chemical equations. Here's how to go about it:
Step 1: Write and balance the chemical equation:
2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
Step 2: Calculate the molar mass of AgNO3:
AgNO3 has a molar mass of 169.87 g/mol (from the periodic table).
Step 3: Calculate the number of moles of AgNO3:
Using the given mass of AgNO3 (50 g) and its molar mass (169.87 g/mol):
moles of AgNO3 = mass / molar mass = 50 g / 169.87 g/mol = 0.294 mol
Step 4: Determine the stoichiometric ratio between AgNO3 and CaCl2:
From the balanced equation, we see that 2 moles of AgNO3 react with 1 mole of CaCl2.
Step 5: Calculate the moles of CaCl2 needed:
Using the stoichiometric ratio, we can calculate the moles of CaCl2 required:
moles of CaCl2 = (moles of AgNO3) / 2 = 0.294 mol / 2 = 0.147 mol
Step 6: Calculate the volume of the CaCl2 solution needed:
We know the concentration of CaCl2 is 1.45 M, which means there are 1.45 moles of CaCl2 per liter of solution.
volume of CaCl2 solution = moles of CaCl2 / concentration of CaCl2
volume = 0.147 mol / 1.45 mol/L = 0.101 L = 101 mL
Therefore, you would need approximately 101 mL of the 1.45 M CaCl2 solution to precipitate all of the silver as AgCl.
To determine the volume of the 1.45M CaCl2 solution needed, we first need to calculate the moles of AgNO3 present in 50 g of AgNO3.
Step 1: Convert grams of AgNO3 to moles.
The molar mass of AgNO3 is 169.87 g/mol (107.87 g/mol for Ag + 14.01 g/mol for N + 3 * 16.00 g/mol for O).
Moles = mass / molar mass
Moles = 50 g / 169.87 g/mol
Moles = 0.294 mol AgNO3
Step 2: Use the balanced equation to determine the stoichiometric ratio between AgNO3 and CaCl2.
From the balanced equation:
2AgNO3 + CaCl2 > 2AgCl + Ca(NO3)2
We can see that the ratio between AgNO3 and CaCl2 is 2:1.
Step 3: Calculate the moles of CaCl2 needed.
Using the ratio found in step 2, we can calculate the moles of CaCl2 needed:
Moles of CaCl2 = 0.294 mol AgNO3 / 2 = 0.147 mol CaCl2
Step 4: Calculate the volume of the 1.45M CaCl2 solution needed.
Molarity (M) = moles / volume (in liters)
Volume (in liters) = moles / Molarity
Volume of 1.45M CaCl2 solution = 0.147 mol CaCl2 / 1.45 mol/L
Volume = 0.101 L or 101 mL
Therefore, you would need approximately 101 mL of the 1.45M CaCl2 solution to precipitate all of the silver as AgCl.