Question

Iron and oxygen form rust. How many moles of rust should be produced if 1
55.321 grams of iron reacted? *

Answer 1

Answer:

1.387 moles

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

4Fe + 3O2 —> 2Fe2O3

Step 2:

Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:

Mass of Fe = 155.321g

Molar Mass of Fe = 56g/mol

Number of mole of Fe =?

Number of mole = Mass/Molar Mass

Number of mole of Fe = 155.321/56

Number of mole of Fe = 2.774 mol

Step 3:

Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:

From the balanced equation above,

4 moles of Fe produced 2 moles of Fe2O3.

Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.

Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction

Answer 2

Answer:

$n = 0.496\,moles$

Explanation:

The stoichometric formula of the chemical process is:

$2Fe + 1.5O_{2} \rightharpoonup Fe_{2}O_{3}$

One mole of rust is formed from 2 moles of iron. Then, the molar ratio of rust to iron is:

$r = \frac{1}{2}$

The quantity of moles of iron is:

$n =\frac{55.321\,\frac{g}{mol} }{55.845\,\frac{g}{mol}}$

$n = 0.991\,moles$

The quantity of moles of rust is:

$n = \frac{1}{2}\cdot(0.991\,moles)$

$n = 0.496\,moles$