Iron can be obtained by reduction of iron oxide ($F{e_3}{O_4}$)with CO according to the reaction:
$F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}$
How many kg of $F{e_3}{O_4}$ should be heated with CO to get 3 Kg of iron?
A. 8.12 Kg
B. 4.14 Kg
C. 6.94 Kg
D. 16.8 Kg

Hint: In the given, iron (II) (III) oxide is reduced with the help of reducing agent carbon monoxide to iron. Three mole of iron is obtained from one mole of iron (II) (III) oxide. Total mass of iron formed from 232 g iron oxide is 168 g.

Complete step by step answer:
The mass of iron is 3kg.
The given reaction is shown below.
$F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}$
In this reaction one mole of iron (II) (III) oxide reacts with four mole of carbon monoxide to form three mole of iron and four mole of carbon dioxide.
In this reaction, iron (II) (III) oxide$F{e_3}{O_4}$is reduced using carbon monoxide to form iron and releases carbon dioxide.
The carbon monoxide is the reducing agent.
The molecular weight of iron Fe is 56 g/mol.
The molecular weight of iron (II) (III) oxide $F{e_3}{O_4}$ is 232 g/mol.
From the balanced equation, it is given that three mole of iron is formed from one mole of iron (II)(III) oxide.
So, $(3 \times 56 = 168g)$ of iron Fe is formed from 232 g iron (II)(III) oxide $F{e_3}{O_4}$.
3 kg is equal to 3000 g.
3000 g of iron Fe will be obtained from $\Rightarrow \dfrac{{232}}{{168}} \times 3000g = 4142.8g$ iron (II) (III) oxide $F{e_3}{O_4}$.
4142.8 g is equal to 4.142 Kg $F{e_3}{O_4}$.
Thus, 4.142 Kg of iron (II) (III) oxide $F{e_3}{O_4}$ should be heated with CO to get 3 Kg of iron.

So, the correct answer is Option B.

Note: The mass is given in kg which needs to be converted into grams as the molecular weight is calculated in terms of gram per mole g/mol. 1 Kg is equal to 1000 g. 1kg = 1000 g.