Question

Equal weights of Zn metal and iodine are mixed together and I₂ is completely converted to ZnI₂. What fraction by weight of the original Zn remains unreacted? (Zn = 65, I = 127)

Options

• 0.34

• 0.74

• 0.84

• unable to predict

Answer 1

0.74

Explanation:

Let x be the weight of Zn an I₂ initially.

By the equation, \[\ce{Zn + I2 -> ZnI2}\]

The initial number of moles of zinc and iodine is `x/65 and x/254` moles respectively.

The number of moles of Zn at the end of the reaction is `(x/65 - x/254)` moles.

∴ Fraction of Zn remained unreacted = `((x/65 - x/254))/(x/65)` = 0.74