If 30.0 mL of 0.150 M CaC12 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgC1 precipitate?

2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2

moles CaCl2 = M x L = ??
moles AgCl produced is 2x moles CaCl2 = ??
g AgCl = moles AgCl x molar mass AgCl.

1290.15grams

To determine the mass of AgCl precipitate formed, we first need to determine the limiting reactant in the reaction between CaCl2 and AgNO3. The limiting reactant is the one that gets completely consumed and determines the maximum amount of product that can be formed.

Step 1: Calculate the moles of CaCl2:
Moles of CaCl2 = volume (in liters) × molarity
Moles of CaCl2 = 0.030 L × 0.150 mol/L
Moles of CaCl2 = 0.0045 mol

Step 2: Calculate the moles of AgNO3:
Moles of AgNO3 = volume (in liters) × molarity
Moles of AgNO3 = 0.015 L × 0.100 mol/L
Moles of AgNO3 = 0.0015 mol

Step 3: Determine the stoichiometry between CaCl2 and AgCl. From the balanced chemical equation, we know that:
CaCl2 + 2AgNO3 → Ca(NO3)2 + 2AgCl

Step 4: Determine the limiting reactant. Since the stoichiometry between CaCl2 and AgCl is 1:2, the limiting reactant will be AgNO3, as it requires twice as many moles as CaCl2 for the reaction.

Step 5: Calculate the moles of AgCl formed using the limiting reactant:
Moles of AgCl = Moles of AgNO3 × 2
Moles of AgCl = 0.0015 mol × 2
Moles of AgCl = 0.0030 mol

Step 6: Calculate the mass of AgCl formed using its molar mass:
Mass of AgCl = Moles of AgCl × molar mass of AgCl
Molar mass of AgCl = Atomic mass of Ag + Atomic mass of Cl
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol
Molar mass of AgCl = 143.32 g/mol

Mass of AgCl = 0.0030 mol × 143.32 g/mol
Mass of AgCl = 0.4299 g

Therefore, the mass of AgCl precipitate formed is approximately 0.430 grams.

To find the mass of the AgCl precipitate formed, we first need to determine the limiting reactant in the reaction between CaCl2 and AgNO3. The limiting reactant is the one that is completely consumed and determines the extent of the reaction.

1. Calculate the number of moles of CaCl2:
Moles of CaCl2 = volume (in liters) × concentration (in moles per liter)
= 0.030 L × 0.150 mol/L

2. Calculate the number of moles of AgNO3:
Moles of AgNO3 = volume (in liters) × concentration (in moles per liter)
= 0.015 L × 0.100 mol/L

3. Use the stoichiometry of the balanced equation to determine the mole ratio between CaCl2 and AgCl. The balanced equation is:
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2

According to the equation, 1 mole of CaCl2 reacts with 2 moles of AgCl. So, to convert moles of CaCl2 to moles of AgCl:
Moles of AgCl = Moles of CaCl2 × (2 mol AgCl / 1 mol CaCl2)

4. Compare the moles of AgCl calculated in step 3 with the moles of AgNO3 from step 2 to determine the limiting reactant. The limiting reactant is the one with fewer moles.

5. Calculate the mass of AgCl precipitate:
Mass of AgCl = Moles of AgCl × molar mass of AgCl

The molar mass of AgCl can be found by adding the atomic masses of silver (Ag) and chlorine (Cl) from the periodic table.

By following these steps, you will be able to calculate the mass in grams of AgCl precipitate formed in the reaction between CaCl2 and AgNO3.