Question

The synthesis of zinc iodide was performed using 1.0582 g of zinc and 2.6987 g of iodine. Determine the limiting reactant and calculate the theoretical yield of zinc iodide in grams.

Answer 1

Zinc iodide can be made by the reaction of Zinc with iodine according to the following equation:                                       Zn+ I2 ------> ZnI2

Have you checked to make sure the equation is balanced correctly?

Find the molecular masses of all species involved. Zn = 65.38 g/mol      I2 = 253.809 g/mol      ZnI2 = 319.22 g/mol

Convert the grams into moles. moles of Zn = g/mm = 1.05820 g/65.38 g/mol = 0.016 moles of zinc on hand. moles of I2 = g/mm = 2.6987 g/253.809 g/mol = 0.010 moles of iodine on hand.

Decision time: Which is the limiting reagent? IF we use all zinc then:                                Zn       = I2                             0.016 mol    x          x = 0.016 moles of iodine are needed.

We don't have 0.016 moles of iodine. We have 0.010 moles of iodine. Therefore we will run out of iodine first. It is the limiting reactant.

Confirmation: If we use all the chlorine then:                                 Zn       =    I2                                    x            0.010 mol       x = 0.010 moles of Zinc are needed.

We have 0.016 moles of Zinc, therefore it is in excess. If it is in excess then the iodine is the limiting reactant.

Use the limiting reactant to cross the ratio bridge and find the moles of ZnI2 that will be produced.                                      I2                =        ZnI2                                   0.010 mol             x                   x = 0.010 moles of ZnI2 are produced

Grams of Zinc iodide are found with g = n * mm = 0.010 mol * 319.22 g/mol = 3.1922 g

Finishing statement: When 1.0582 grams of Zinc and 2.6987 grams of iodine are reacted according to the above equation, the iodine is the limiting reactant and the maximum yield of Zinc iodide is 0.010 moles or 3.1922 grams.