Question

56 g of Fe reacts with 32 g of oxygen to produces Fe 2O3. The moles of Fe2O3 produced will be ​

Answer 1

Answer:

Iron (III) oxide is formed according to the following balanced chemical equation: 4Fe(g) + 3O2(g)→ 2Fe2O3(g) The reaction of 8.0 moles of solid iro

Answer 2

Explanation:

this question is based on stoichiometry

In 56g of fe ,no of moles = given mass / molar mass=56/56= 1 mole of Fe

In 32g of O2,no of moles=32/32 =1 mole of O2

We want to balance the equation

4Fe + 3O2 --> 2Fe2O3

(4moles) (3 moles) 2 moles

therefore 4moles of Fe requires 3 moles of O2

so 1 mole of Fe requires 3/4 moles of O2

according to given

1 mole if Fe requires 1 mole of O2

so ,O2 is the limiting reagent

Now we want to compare LHS and RHS,

3 moles of 02 will produce 2 moles Fe2O3

so 1 mole of O2 will produce 2/3moles of Fe2O3

this is the right answer