To determine the mass of silver chloride precipitated from a solution containing calcium chloride, we need to calculate the stoichiometry of the reaction.
The balanced chemical equation for the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3) is:
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2
From the equation, we can see that for every 1 mole of calcium chloride, 2 moles of silver chloride are produced.
Step 1: Calculate the number of moles of calcium chloride (CaCl2):
Given: Mass of calcium chloride = 1.50 g
Molar mass of CaCl2 = 40.08 g/mol + (2 × 35.45 g/mol) = 110.98 g/mol
Number of moles = Mass / Molar mass = 1.50 g / 110.98 g/mol ≈ 0.0135 mol
Step 2: Calculate the mass of silver chloride (AgCl) produced:
From the stoichiometry of the reaction, we know that 2 moles of AgCl are produced for every 1 mole of CaCl2.
So, the molar ratio of AgCl to CaCl2 is 2:1.
Mass of AgCl = Number of moles of CaCl2 × Molar mass of AgCl
Number of moles of AgCl = 2 × Number of moles of CaCl2 = 2 × 0.0135 mol = 0.0270 mol
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = 0.0270 mol × 107.87 g/mol ≈ 2.92 g
Therefore, approximately 2.92 grams of silver chloride (AgCl) will precipitate from the solution containing 1.50 grams of calcium chloride (CaCl2).