Chemistry 101 HW

Question

Part A

What mass of silver chloride can be produced from 1.88 L of a 0.284 M solution of silver nitrate?

Part B

The reaction described in Part A required 3.68 L of calcium chloride. What is the concentration of this calcium chloride solution?

Answer 1

A.). AgNO₃ + Cl^- => AgCl

moles AgNO₃ = 1.88 L x 0.284 mol/L = 0.534 mol

moles AgCl = 0.534 mol

mass AgCl = 0.534 mol x 143 g/mol = 76.4 g

B.). CaCl₂+ 2AgNO₃ ==> 2AgCl + CaNO₃

moles AgNO3 = 0.534 mole (from part A)

moles CaCl₂ needed = 0.534 mol AgNO₃ x 1 mol CaCl2/2mol AgNO3 = 0.267 mol CaCl2

3.68 L (x mol/L) = 0.267 moles = 0.0726 M