2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2 When 13.5 moles of CuCl2 reacts with 19.86 moles of KI, how many moles of I2 are formed?

6.75 moles

Explanation:

This is a simple stoichiometry proboe. that I would set up like this:

(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)

That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.

Hope this helps :)

answer: gold foil experiment - he shot alpha particles at a gold foil

explanation: