In the preparation of iron from haematite $\left( {F{e_2}{O_3}} \right)$ by the reaction with carbon as
$F{e_2}{O_3} + C \to Fe + C{O_2}$
How much 80% pure iron could be produced from 120 kg of 90% pure $F{e_2}{O_3}$?
A.94.5kg
B.60.48kg
C.116.66kg
D.120kg
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Hint: We have to calculate the mass of 80% pure iron that could be produced from $F{e_2}{O_3}$ using the number of moles of $F{e_2}{O_3}$ and the atomic mass along with mass percentage of pure iron.
Complete step by step answer:
Given data contains,
The percentage of pure iron is 80%.
The percentage of pure iron oxide is 90%.
The mass of iron oxide is 120kg.
Iron oxide that is $F{e_2}{O_3}$ on reaction with carbon forms metallic iron and carbon dioxide. The chemical reaction is written as,
$F{e_2}{O_3} + C \to Fe + C{O_2}$
We can see that the given reaction is unbalanced. Let us first balance the reaction.
On the reactant side, we need 2 moles of $F{e_2}{O_3}$ and three moles of carbon.
On the product side, we need 4 moles of iron and three moles of carbon dioxide.
We can write the balanced reaction as,
$2F{e_2}{O_3} + 3C \to 4Fe + 3C{O_2}$
Let us now calculate the moles of $F{e_2}{O_3}$ using the given mass of $F{e_2}{O_3}$ , the molar mass and percentage of pure $F{e_2}{O_3}$
The molar mass of iron oxide is 160g/mol.
We can write the formula to calculate the number of moles as,
Number of moles$ = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}} \times {\text{percentage}}s$
Let us substitute the values of mass, percentage of pure $F{e_2}{O_3}$ and molar mass to calculate the number of moles.
Number of moles$ = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}} \times {\text{percentage}}$
Number of moles=$\dfrac{{120 \times 1000}}{{160}} \times \dfrac{{90}}{{100}}$
Number of moles=$\dfrac{{120 \times 100 \times 9}}{{160}}$
Let us now calculate the mass of 80% pure iron reduced.
The atomic mass of pure iron is 56g.
The oxidation state of pure iron is +2.
Mass of 80% pure iron=$\dfrac{{120 \times 100 \times 9}}{{160}} \times \dfrac{{2 \times 56}}{{\dfrac{{80}}{{100}}}}$
Mass of 80% pure iron=$\dfrac{{120 \times 100 \times 9 \times 112 \times 100}}{{160 \times 80}}$
Mass of 80% pure iron=$94500g$
So, we have determined the Mass of 80% pure iron as $94500g$. We have to convert the grams into kilograms. So divide the obtained grams by 1000.
$kg = 94500g \times \dfrac{{1kg}}{{1000g}}$
$kg = 94.5kg$
Mass of 80% pure iron=$94.5kg$
The mass of 80% pure iron is $94.5kg$.
Therefore, the option (A) is correct.
Note:
We must remember that the iron ores is the fourth most abundant element in the crust of the earth. Iron reacts with oxygen and water to produce brown to black hydrated iron oxides, generally called as rust. Iron forms compounds in ranges of oxidation states, −2 to +7.
Complete step by step answer:
Given data contains,
The percentage of pure iron is 80%.
The percentage of pure iron oxide is 90%.
The mass of iron oxide is 120kg.
Iron oxide that is $F{e_2}{O_3}$ on reaction with carbon forms metallic iron and carbon dioxide. The chemical reaction is written as,
$F{e_2}{O_3} + C \to Fe + C{O_2}$
We can see that the given reaction is unbalanced. Let us first balance the reaction.
On the reactant side, we need 2 moles of $F{e_2}{O_3}$ and three moles of carbon.
On the product side, we need 4 moles of iron and three moles of carbon dioxide.
We can write the balanced reaction as,
$2F{e_2}{O_3} + 3C \to 4Fe + 3C{O_2}$
Let us now calculate the moles of $F{e_2}{O_3}$ using the given mass of $F{e_2}{O_3}$ , the molar mass and percentage of pure $F{e_2}{O_3}$
The molar mass of iron oxide is 160g/mol.
We can write the formula to calculate the number of moles as,
Number of moles$ = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}} \times {\text{percentage}}s$
Let us substitute the values of mass, percentage of pure $F{e_2}{O_3}$ and molar mass to calculate the number of moles.
Number of moles$ = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}} \times {\text{percentage}}$
Number of moles=$\dfrac{{120 \times 1000}}{{160}} \times \dfrac{{90}}{{100}}$
Number of moles=$\dfrac{{120 \times 100 \times 9}}{{160}}$
Let us now calculate the mass of 80% pure iron reduced.
The atomic mass of pure iron is 56g.
The oxidation state of pure iron is +2.
Mass of 80% pure iron=$\dfrac{{120 \times 100 \times 9}}{{160}} \times \dfrac{{2 \times 56}}{{\dfrac{{80}}{{100}}}}$
Mass of 80% pure iron=$\dfrac{{120 \times 100 \times 9 \times 112 \times 100}}{{160 \times 80}}$
Mass of 80% pure iron=$94500g$
So, we have determined the Mass of 80% pure iron as $94500g$. We have to convert the grams into kilograms. So divide the obtained grams by 1000.
$kg = 94500g \times \dfrac{{1kg}}{{1000g}}$
$kg = 94.5kg$
Mass of 80% pure iron=$94.5kg$
The mass of 80% pure iron is $94.5kg$.
Therefore, the option (A) is correct.
Note:
We must remember that the iron ores is the fourth most abundant element in the crust of the earth. Iron reacts with oxygen and water to produce brown to black hydrated iron oxides, generally called as rust. Iron forms compounds in ranges of oxidation states, −2 to +7.