How many grams of Fe2O3 react with excess Al to make 475 g Fe?no

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Atomic weight of Iron (Fe) is 55.874 and that of oxygen (O) is 15.9994.

As each Fe2O3 molecule contains 2 atoms of iron and 3 atoms of oxygen, the proportion of mass of iron in Fe2O3

(Mass of iron)/(Mass of Fe2O3)

= [2x(Atomic weight of iron)]/[2x(Atomic weight of iron) +3x(Atomic weight of Oxygen)]

= (2x55.847)/(2x55.847 + 3x15.9994)

= 111.694/(111.694 + 47.9982) = 0.6517

Therefore weight of Fe2O3 required to yield 475 g of Fe

= 475/0.651709

= 728.8529 g

Answer:

728.8529 g of Fe2O3 will react with excess aluminium (Al) to produce 475 g of iron (Fe).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To determine how many grams of F2 O3 react with Al to make 475 g of Fe.

Solution:

We know Fe2 O3 +2 Al  --> 2Fe + A2 O3. So,

(One mole of F2 O3) + (2 mole of AL) = (2mole of Fe) + (one mole of Al2 O3).

So, by mass ,

(55.85g*2+16g*3) + (2*26.98g) = 2*58.85g Fe +..... ..  Or

(159.7g of F2O3) + 53.96 g of AL = 111.7 g of Fe.+..... So to get 475 g of Fe, the right side we have to mutiply 475/111.7.

This requires a multification by 475/117.7 both side of the equation.

Therefore, to produce 475g of Fe = (111.7)(475/111.7) Fe  requires  165.7*475/111.7 gram  of  Fe2 O3 to react with 2*26*98*(475/111.7)g of Al.

So the required F2 O3 = 159.7*(475/111.7) = 679.1181737gram

The required Aluminium = 2*26.98*475/111.7 = 229.4628469gram.

 

 

 

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