Question

Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s)is mixed with 11.5 L of O2(g) at 2.66 atm and 298 K.(a) Calculate the number of moles of each of the following before the reaction occurs.(i) Fe(s)(ii) O2(g)(b) Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support youranswer with calculations.(c) Calculate the number of moles of Fe2O3 produced when the reaction proceeds tocompletion.

Answer 1

a) 1,34 Fe(s) moles and 1,25 O₂(g) moles.

b) Limiting reagent is Fe(s).

c) 0,67 Fe₂O₃ moles

Step-by-step explanation:

a) The first thing we should know is how many moles are of each reagent, thus:

• 75,0 Fe(s) grams ×
  `(1 mole)/(55,845 g)` = 1,34 Fe(s) moles

-Using atomic mass of Iron-

• To obtain O₂ moles it is necessary to use ideal gas law:

`(PV)/(RT)` = n

Where:

P is pressure: 2,66 atm

V is volume: 11,5 L

R is gas constant: 0,082
`(atm.L)/(mol.K)`

T is temperature: 298 K

n are unknowed moles.

Thus, moles are: 1,25 O₂ moles

b) The global reaction is:

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃

So, four Fe(s) moles reacts with three O₂(g) moles. It means that for a complete reaction of 1,34 Fe(s) moles you need:

1,34 Fe(s) moles ×
`(3 O_(2moles) )/(4 Fe moles)` = 1,01 O₂ moles

For a complete reaction of this iron you need just 1,01 O₂ moles but you have 1,25 O₂ moles. Thus, limiting reagent is Fe(s).

c) Now, the produced Fe₂O₃ moles are calculated with the limiting reagent moles, knowing that 4 Fe(s) moles produce 2 Fe₂O₃ moles -global reaction-, thus:

1,34 Fe(s) moles ×
`(2 Fe_(2)O_(3)moles  )/(4Fe moles)` =

0,67 Fe₂O₃ moles

I hope it helps!