Suppose that 75 mL of 5.00 M iron (III) chloride is combined with 81 mL of 4.66 M calcium chloride. What is the concentration of Cl- ion in the resulting solution?

The concentration of chloride ion in the resulting solution is 12.04 M

Explanation:

To calculate the molarity of solution, we use the equation:

       .......(1)

Molarity of solution = 5.00 M

Volume of solution = 75.00 mL

Putting values in equation 1, we get:

1 mole of iron (III) chloride produces 1 mole of ions and 3 moles of ions

Moles of chloride ions in iron (III) chloride solution = (3 × 0.375) = 1.125 moles

Molarity of solution = 4.66 M

Volume of solution = 81 mL

Putting values in equation 1, we get:

1 mole of calcium chloride produces 1 mole of ions and 2 moles of ions

Moles of chloride ions in calcium chloride solution = (2 × 0.377) = 0.754 moles

Now, calculating the chloride ions in the solution by using equation 1, we get:

Total moles of chloride ions = [1.125 + 0.754] =  1.879 moles

Total volume of base solution = [75 + 81] = 156 mL

Putting values in equation 1, we get:

Hence, the concentration of chloride ion in the resulting solution is 12.04 M

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

catya that's not even a b or c