A piece of iron weighing 251.3 g was heated and reacted with oxygen. The mass of the product (iron (II) oxide) was 395.3 g. 1. How many moles of iron were used to produce the iron oxide? a. 3.0...

A piece of iron weighing 251.3 g was heated and reacted with oxygen. The mass of the product (iron (II) oxide) was 395.3 g.

1. How many moles of iron were used to produce the iron oxide?

a. 3.0 moles

b. 3.5 moles

c. 4.0 moles

d. 4.5 moles

e. 5.0 moles

2. How many moles of oxygen were used to produce the iron oxide?

a. 9.0 moles

b. 5.0 moles

c. 6.5 moles

d. 4.0 moles

e. 7.0 moles

Asked on by oliviawee

2 Answers | Add Yours

gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

The chemical reaction for this case:

`Fe + 1/2 O_2 -> FeO`

Molar mass of Fe = 55.85 gm/mole

Molar mass of oxygen = 32 gm/mole

Molar mass of Fe(II) oxide = 55.85+16 = 71.85 gm/mole

Initial mass of iron = 251.3 gm = 251.3/55.85 moles = 4.5 moles

Mass of the product (FeO) = 395.3 gm = 395.3/71.85 moles = 5.5 moles

From stoichiometry,

1 mole of iron ~ 1 mole of Fe(II)oxide ~ 1/2 mole oxygen

Please check the question, there seems to be something wrong with the number.

Let me explain.

as per stoichiometry,

1 mole of iron will produce 1 mole of iron (II) oxide. However, if the numbers were true, in the present case 4.5 mole of iron is generating 5.5 mole of iron oxide. 

Also, if we calculate the amount of oxygen consumed, we get different answers when we use reactant data and product data.

for ex., 1 mole iron uses 0.5 mole oxygen

so, 4.5 mole iron will use 4.5 x 0.5 = 2.25 moles oxygen.

and 1 mole of product is generated by 0.5 mole oxygen

so, 5.5 mole of product should be generated from 5.5 x 0.5 = 2.75 moles oxygen.

Interestingly, none of these are the available answer options.

Using 4.5 moles iron, we can generate only 4.5 moles of iron oxide and in the process will consume 2.25 moles of oxygen. 

Please check the question. 

hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

The balanced reaction is :-

Fe(s) + (1/2)O2(g)  ---------> FeO(s)

Molar mass of Fe = 56 g/mole 

Molar mass of O2 = 32 g/mole

Molar mass of FeO = 72 g/mole

Now, moles of Fe present in 251.3 g of it = mass/molar mass = 251.3/56 = 4.4875

moles of FeO present in 395.3 g of it = mass/molar mass = 395.3/72 = 5.49

Now, As per the balanced reaction,

1 mole of Fe on complete reaction with excess O2(or 0.5 moles of O2) produces 1 mole of FeO

But as per the question given, the moles of Fe present is lesser than the moles of FeO produced.

So, please re-check the question and make the necessary changes.

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