What mass of precipitate is formed when 0.50 L of 2.50 mol/L aqueous CaCl2 with 1100 mL of 3.50 mol/L aqueous silver nitrate?
To determine the mass of precipitate formed in this reaction, we need to calculate the number of moles of the limiting reactant, which will be completely consumed. The limiting reactant is the one that is present in a smaller quantity, and it determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of CaCl2:
Moles of CaCl2 = concentration (mol/L) × volume (L)
= 2.50 mol/L × 0.50 L
= 1.25 mol
Next, we'll calculate the number of moles of AgNO3:
Moles of AgNO3 = concentration (mol/L) × volume (L)
= 3.50 mol/L × 1.10 L
= 3.85 mol
Now, we need to determine the stoichiometric ratio between the reactants in the balanced chemical equation:
CaCl2 + 2AgNO3 -> Ca(NO3)2 + 2AgCl
From the equation, we can see that one mole of CaCl2 reacts with two moles of AgNO3 to produce one mole of precipitate (AgCl).
The mole ratio tells us that 1 mole of CaCl2 produces 1 mole of AgCl. Therefore, the number of moles of AgCl formed will be equal to the number of moles of CaCl2.
Now, we can calculate the mass of AgCl (precipitate) formed:
Mass of AgCl = moles of AgCl × molar mass of AgCl
First, let's calculate the number of moles of AgCl:
Moles of AgCl = moles of CaCl2 = 1.25 mol
The molar mass of AgCl is calculated by adding the atomic masses of silver (Ag) and chlorine (Cl):
Molar mass of AgCl = (atomic mass of Ag) + (atomic mass of Cl)
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Now we can calculate the mass of AgCl:
Mass of AgCl = moles of AgCl × molar mass of AgCl
= 1.25 mol × 143.32 g/mol
= 179.15 g
Therefore, the mass of precipitate (AgCl) formed is 179.15 grams.