Answer :
Answer:
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!