Answer:
There are 0.100 moles chlorine ions (Cl-) involved
Step-by-step explanation:
Step 1: Data given
Volume of CaCl2 = 25.0 mL = 0.025 L
Molarity of CaCl2 = 2.00 M
Step 2: The balanced equation
CaCl2 (aq) + 2AgNO3 (aq) ⇆ Ca(NO3)2 (aq) + 2AgCI (s)
CaCl2 → Ca^2+ + 2Cl-
Step 3: Calculate moles CaCl2
Moles CaCl2 = molarity CaCl* volume
Moles CaCl2 = 2.00 M * 0.025 L
Moles CaCl2 = 0.050 moles
Step 4: Calculate moles chloride ions
CaCl2 → Ca^2+ + 2Cl-
For 1 mol CaCl2 we have 1 mol Ca^2+ and 2 moles Cl-
Moles Cl- ions = 2*0.050 moles
Moles Cl- ions = 0.100 moles
There are 0.100 moles chlorine ions (Cl-) involved