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

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⇒ 2

From the given weight of iron to be used for the production of , 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.

indium, and thallium have three electrons i think

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.