If an excess of silver nitrate is combined with 25.0 mL of calcium chloride and produces 2.90 g of precipitate, what is the molarity of calcium chloride that was used?
CaCl2+AgNO3>>2AgCl + ....
Ok, you first figure the moles of silver chloride form 2.9grams.
Then you had half that number of moles of calcium chloride in solutions, so
MolarityCaCl2=molesCaCl2/.025
To find the molarity of calcium chloride, we need to first convert the mass of the precipitate to moles, then divide by the volume of calcium chloride used in liters.
1. Calculate moles of precipitate:
To do this, we need to find the molar mass of the precipitate. The precipitate is formed by the reaction of silver nitrate (AgNO3) and calcium chloride (CaCl2). The balanced equation is:
2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2
The molar mass of AgCl is 143.32 g/mol. Therefore, the moles of AgCl produced can be calculated as follows:
moles of AgCl = mass of AgCl / molar mass of AgCl
= 2.90 g / 143.32 g/mol
2. Calculate volume of calcium chloride in liters:
The volume of calcium chloride given in the problem is 25.0 mL. We need to convert this to liters by dividing by 1000:
volume of CaCl2 in liters = 25.0 mL / 1000 mL/L
3. Calculate molarity of calcium chloride:
Finally, we can calculate the molarity of calcium chloride using the equation:
Molarity (M) = moles / volume in liters
Molarity of CaCl2 = moles of CaCl2 / volume of CaCl2 in liters
= moles of AgCl / volume of CaCl2 in liters (Since 1 mole of CaCl2 produces 2 moles of AgCl)
Now, you can substitute the values into the equation to find the molarity of calcium chloride.
To find the molarity of calcium chloride, we need to calculate the number of moles of calcium chloride and the volume of the solution it was dissolved in.
First, let's find the number of moles of silver nitrate used. We can do this by dividing the mass of the precipitate by the molar mass of silver nitrate (AgNO3). The molar mass of AgNO3 is:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in AgNO3)
So, the molar mass of AgNO3 is 107.87 + 14.01 + (16.00 * 3) = 169.87 g/mol.
Next, we can calculate the number of moles of AgNO3:
moles of AgNO3 = mass of precipitate / molar mass of AgNO3
moles of AgNO3 = 2.90 g / 169.87 g/mol
moles of AgNO3 ≈ 0.01708 mol
Since calcium chloride (CaCl2) reacts in a 1:2 ratio with silver nitrate in the precipitation reaction, the number of moles of calcium chloride is twice the number of moles of silver nitrate:
moles of CaCl2 = 2 * moles of AgNO3
moles of CaCl2 ≈ 2 * 0.01708 mol
moles of CaCl2 ≈ 0.03416 mol
Now let's calculate the volume of the calcium chloride solution. We are given that the volume of calcium chloride solution is 25.0 mL, which is equivalent to 0.0250 L.
Finally, we can find the molarity (M) of calcium chloride:
Molarity (M) = moles of solute / volume of solution
Molarity (M) = 0.03416 mol / 0.0250 L
Molarity (M) ≈ 1.3664 mol/L
Therefore, the molarity of the calcium chloride solution is approximately 1.37 mol/L.