Question

How do I calculate the mass of zinc and iodine that are consumed to produce zinc iodide?

Answer 1

In a chemical reaction, the mass of a substance that is consumed refers to one or more reactants. In the case of your question, the reactants are zinc and iodine, so you are being asked to determine the mass of zinc and the mass of iodine that were consumed to form zinc iodide. The balanced chemical equation for this synthesis reaction between zinc and iodine is:

`"Zn"` + `"I"_2` `rarr` `"ZnI"_2`

In order to determine the mass of zinc consumed (reacted), you must know the mass of iodine that was consumed (reacted), or the mass of zinc iodide that was produced. In order to determine the mass of iodine consumed, you must know the mass of zinc that was consumed, or the mass of zinc iodide produced.

Example 1

How many grams of `"Zn"` must be consumed to react with `"12.52g I"_2` to produce `"ZnCl"_2`?

Solution

1. Determine moles `"I"_2"` by dividing the given mass by its molar mass. I prefer to divide by multiplying by the inverse of the molar mass, `"mol/g"`.

2. Calculate moles `"Zn"` by multiplying moles `"I"_2` by the mole ratio between `"I"_2"` and `"Zn"` in the balanced equation, with `"Zn"` in the numerator.

3. Determine the mass of the `"Zn"` required to react with `"12.5 g I"_2"` by multiplying the moles `"Zn"` by it molar mass.

`12.5color(red)cancel(color(black)("g I"_2))xx(1color(red)cancel(color(black)("mol I"_2)))/(253.808color(red)cancel(color(black)("g I"_2")))` `xx` `(1color(red)cancel(color(black)("mol Zn")))/(1color(red)cancel(color(black)("mol I"_2")))` `xx` `(65.38"g Zn")/(1color(red)cancel(color(black)("mol Zn")))="3.22 g Zn"`

`"3.22g Zn"` are consumed when reacted with `"12.5g I"_2"`

Example 2

How many grams of `"Zn"` must be consumed to react with an excess of `"I"_2"` to produce `"15.7 g ZnI"_2"`?

Follow the same procedure, substituting `"ZnI"_2"` for `"I"_2"`, and `"319.189 g/mol"` for the molar mass in the first step. Use the mole ratio between `"Zn"` and `"ZnI"_2"`, with `"Zn"` in the numerator. Multiply by the molar mass of `"Zn"`.

`15.7color(red)cancel(color(black)("g ZnI"_2))xx(1color(red)cancel(color(black)("mol ZnI"_2)))/(319.189color(red)cancel(color(black)("g ZnI"_2)))` `xx` `(1color(red)cancel(color(black)("mol Zn")))/(1color(red)cancel(color(black)("mol ZnI"_2)))` `xx` `(65.38"g Zn")/(1color(red)cancel(color(black)("mol Zn")))="3.22 g Zn"`

`"3.22 g Zn"` must be consumed to react with excess `"I"_2"` to produce `"15.7 g ZnI"_2"`.