How many moles of Fe are needed for the reaction of 12.0
moles of O2?

The balanced reaction equation is:

4

F

e

(

s

)

+

3

O

2

(

g

)

→

2

F

e

2

O

3

(

s

)

Initial moles of oxygen reactant (given) =  

12.0

m

o

l

According to the balanced equation the iron and oxygen reactants combine in a 4:3 molar ratio. Therefore for complete consumption of both reactants we require:

Initial moles of iron reactant =  

4

3

×

12.0

m

o

l

=

16.0

m

o

l

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.