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
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
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.
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.