Question

If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form

Answer 1

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3$O_{2}$⇒ 2 $Fe{2}$$O{3}$

From the given weight of iron to be used for the production of $Fe{2}$$O{3}$, number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

  Putting the values in equation  

Number of moles =  10 gm  ÷ 55.845 gm/mole

                               =  0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

 So,  2÷ 4 = x ÷ 0.179

     2/4 = x/ 0.179

    2 × 0.179 = 4x

     2 × 0.179 / 4 = x

  x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

            =  159.69 grams ×  0.0895

             = 14.292 grams of Fe2O3 formed.