Question

how many grams of AI are required to pteparr 558g of fe by the teduction of fe203 with AI at high temperature? how many moles of fe203 will be required in the reaction?​

Answer 1

Answer:

Equation for the reaction (thermite process):

Fe2O3 + 2Al = Al2O3 + 2Fe

Molar mass:

Aluminium - 26.98 g

Iron - 55.85 g

Iron(III) oxide, Fe2O3 - (55.85 X 2) + (16 X 3) = 111.70 + 48 = 159.7 g

Aluminium oxide, Al2O3 - (26.98 x 2) + (16 x 3 ) = 53.96 + 48 = 101.96 g

(1) To find the mass of aluminium required:

As per the equation, to produce 2 moles of iron, 2 moles of aluminium is required.

So, to get 1 mole of iron ((55.85 g), 1 mole of aluminium (26.98 g) will be required.

Amount of aluminium needed to get 558 g of iron = (558 x 26.98)/55.85 = 269.36 g (answer)

(2) To find the number of moles of iron(III) oxide:

Mass of iron to be produced = 558 g = 558/55.85 = 9.99 moles

As per the equation, 1 mole of Fe2O3 is required to get 2 moles of iron.

So, number of moles of Fe2O3 required to get 9.99 moles of iron = (9.99 x 1)/2 = 4.995 = 5 moles (answer).

Anwers:

(1) Mass of aluminium required = 269.36 g

(2) Number of moles of Fe2O3 required = 5 moles