If 60.0 mL of 0.150 M CaCl2 is added to 50.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?
Write the balanced equation.
Convert CaCl2 to mols. #mols = M x L.
Convert AgNO3 to mols.
Determine the limiting reagent (from the moles that is easily done) but note that the reaction is not a 1:1 reaction.
Convert moles AgCl to grams. Post your work if you need further assistance.
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To find the mass of AgCl precipitate formed, we need to determine the limiting reactant and then use stoichiometry to calculate the mass.
First, let's find the number of moles of each reactant:
Moles of CaCl2 = volume (in liters) x molarity
Moles of CaCl2 = 0.060 L x 0.150 mol/L
Moles of CaCl2 = 0.009 mol
Moles of AgNO3 = volume (in liters) x molarity
Moles of AgNO3 = 0.050 L x 0.100 mol/L
Moles of AgNO3 = 0.005 mol
Now, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we compare the molar ratios of the reactants based on the balanced equation. The balanced equation for the reaction between CaCl2 and AgNO3 is:
CaCl2 + 2AgNO3 -> Ca(NO3)2 + 2AgCl
The stoichiometric ratio between CaCl2 and AgNO3 is 1:2. So, for every 1 mole of CaCl2, we need 2 moles of AgNO3.
Let's calculate the number of moles of AgNO3 needed to react with the moles of CaCl2 available:
Moles of AgNO3 needed = 2 x moles of CaCl2
Moles of AgNO3 needed = 2 x 0.009 mol
Moles of AgNO3 needed = 0.018 mol
Since we have only 0.005 mol of AgNO3, which is less than the required amount of 0.018 mol, AgNO3 is the limiting reactant.
Now, we can calculate the moles of AgCl precipitate formed using stoichiometry:
Moles of AgCl = moles of limiting reactant (AgNO3) x stoichiometric ratio (from balanced equation)
Moles of AgCl = 0.005 mol x 2
Moles of AgCl = 0.01 mol
Finally, we can calculate the mass of AgCl precipitate:
Mass of AgCl = moles of AgCl x molar mass of AgCl
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = 0.01 mol x 107.87 g/mol
Mass of AgCl = 1.08 g
Therefore, the mass of AgCl precipitate formed is 1.08 grams.