Question

How many moles of Fe2O3 will be produced from 27.0g of Fe assuming O2 is available in excess

Answer 1

0.24

Step-by-step explanation:

Answer 2

0.24 moles of Fe2O3 will be produced from 27.0g of Fe assuming O2 is available in excess.

Step-by-step explanation:

Data given:

mass of Fe = 27 grams

mass of oxygen = excess

moles of
`Fe_(2)`
`O_(3)` = ?

Atomic mass of Fe= 55.48 grams/mole

atomic mass of Fe2O3 = 159.69 grams/mole

moles of Fe given =0.48 moles

number of moles =
`(mass)/(atomic mass of 1 mole)`

=
`(27)/(55.48)`

= 0.48 moles

Balanced chemical equation for the reaction is:

4 Fe + 3
`O_(2)` ⇒ 2
`Fe_(2)`
`O_(3)`

4 moles of Fe gives 2 mole of Fe2O3

So, 0.48 moles will give x mole of Fe2O3

`(2)/(4)` =
`(x)/(0.48)`

4x = 2 x 0.48

x =
`(2 X 0.48)/(4)`

x = 0.24 moles

so 0.24 moles of
`Fe_(2)`
`O_(3)` is formed from 27 grams of Fe.