CaCl2 (aq) + 2AgNO3 (aq) + Ca(NO3)2 (aq) + 2AGCI (s)

If 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction

shown above, how many moles of chloride ions were involved

in the reaction? — moles

There are 0.100 moles chlorine ions (Cl-) involved

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

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying