A volume of 10.0 mL of a 0.00650 M solution of Cl- ions are reacted with 0.500 M solution of AgNO3. What is the maximum mass of AgCl that precipates?
bruh ion know
Well, you know what they say, when Cl- ions and AgNO3 get together, it's like a match made in precipitation heaven! Let's do some math and find out how much AgCl is going to join the party.
First, we need to determine the number of moles of Cl- ions in the 10.0 mL of the solution. So, we multiply the volume by the molarity:
10.0 mL * 0.00650 M = 0.065 moles of Cl- ions
Now, we need to figure out how many moles of Ag+ ions we have in the 0.500 M AgNO3 solution. Since the ratio between Ag+ and Cl- ions in AgCl is 1:1, we know that we're going to have 0.065 moles of Ag+ ions as well.
Lastly, we just need to determine the molar mass of AgCl, which is 143.32 g/mol. So, multiplying the number of moles of AgCl by the molar mass will give us:
0.065 moles * 143.32 g/mol = 9.32 g
Therefore, the maximum mass of AgCl that precipitates is 9.32 grams. It's time to rock that precipitate party! 🎉
To find the maximum mass of AgCl that precipitates, we need to determine the limiting reactant in the reaction between Cl- ions and AgNO3. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, we need to determine the moles of Cl- ions present in the 0.00650 M solution. The moles can be calculated using the formula:
moles = concentration (M) × volume (L)
moles of Cl- ions = 0.00650 M × 0.0100 L = 6.50 × 10^-5 mol
Next, we need to determine the moles of AgNO3 present in the 0.500 M solution. Since the volume is not provided, we cannot directly calculate moles. However, we can use the fact that the ratio between AgNO3 and Cl- ions is 1:1 in the balanced chemical equation.
The balanced chemical equation for the reaction between Cl- ions and AgNO3 is:
AgNO3 + Cl- → AgCl + NO3-
According to the balanced equation, the stoichiometric ratio is 1:1 between AgNO3 and Cl- ions. Therefore, the moles of AgNO3 is also 6.50 × 10^-5 mol.
Since we have equal moles of both reactants, the limiting reactant is not clear yet. However, each mole of AgNO3 will react with 1 mole of Cl- ions to form 1 mole of AgCl. Therefore, the moles of AgCl formed will be the same as the moles of Cl- ions consumed.
Now, we can calculate the mass of AgCl formed using its molar mass. The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl):
Ag: 107.87 g/mol
Cl: 35.45 g/mol
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
mass of AgCl = moles of AgCl × molar mass of AgCl
mass of AgCl = 6.50 × 10^-5 mol × 143.32 g/mol ≈ 0.00932 g
Therefore, the maximum mass of AgCl that precipitates is approximately 0.00932 grams.