Zinc and iodine react to form zinc(II) iodine (the reactants and the product are all solids at room temperature).

a. Write a balanced equation for this reaction.
Zn + I2 = ZnI2

b. Suppose that 50.0 g of zinc and iodine are used to form zinc(II) iodine.
i. Assume that the reaction goes to completion, which element will be totally consumed in the formation of the zinc(II) iodine?
ii. What is the limiting reactant?
iii. How many grams of zinc(II) iodine can be produced?
iv. How many grams of the excess element remain un-reacted?

((Sorry this is so much. Thanks to anyone who helps.))

Question

Zinc and iodine react to form zinc(II) iodine (the reactants and the product are all solids at room temperature).

a. Write a balanced equation for this reaction.
Zn + I2 = ZnI2

b. Suppose that 50.0 g of zinc and iodine are used to form zinc(II) iodine.
i. Assume that the reaction goes to completion, which element will be totally consumed in the formation of the zinc(II) iodine?
ii. What is the limiting reactant?
iii. How many grams of zinc(II) iodine can be produced?
iv. How many grams of the excess element remain un-reacted?

((Sorry this is so much. Thanks to anyone who helps.))

wolfe8 · Answer

b. Find the number of moles of Zn and I2

anonymous · Answer

Zn = .765 moles
I = .364 moles

wolfe8 · Answer

Now, since the ratio is 1:1, the reactants will be used up at the same rate. Which one do you think will finish first?

anonymous · Answer

Iodine? There is less moles of it.

wolfe8 · Answer

Yup! When I2 runs out, the reaction has to stop because there is nothing for Zn to react with. You should be able to answer i and ii now. For iii, since the reaction ends when I2 is finished, the amount of product will depend on the amount of I2.