If the initial concentrations are {AgNO3} = 0.534 M and {CaCl2}= 0.194 M, how many grams of AgCl will precipitate out of solution?

55,6g of AgCl

Explanation:

The solution contains 1L

For the reaction:

2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + CaNO₃(aq)

For a complete reaction of 0,194 moles of CaCl₂ you need:

0,194 moles of CaCl₂ × (2mol AgNO₃ / 1mol CaCl₂) = 0,388 moles of AgNO₃.

As you have 0,534moles, CaCl₂ is the limiting reactant. That means moles of AgCl that will precipitate are:

0,194 moles of CaCl₂ × (2mol AgCl / 1mol CaCl₂) = 0,388 moles of AgCl

In grams (MW AgCl: 143,32g/mol):

0,388 mol × (143,32g / 1mol) = 55,6g of AgCl

I hope it helps!