In the preparation of iron from haematite - -Fe-2O-3- by the reaction with carbon as given below- Fe-2O-3-C - Fe-CO-2- How much - 80 - pure iron could be produced from - 120 kg- of - 90 - pure - Fe-2O-3- 94-5 kg- 60-48 kg- 116-66 kg- 120 kg-

The correct option is A 94.5 kg Balanced reaction is nbsp; 2Fe_2O_3 +3C → 4Fe+3CO_2 .Number of moles of Fe_2O_3 is ( 120 × 10002 × 56+48 ...

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Standard XII

Chemistry

Synthesis by Reduction of Nitriles

In the prepar...

Question

In the preparation of iron from haematite $(F e_{2} O_{3})$ by the reaction with carbon as given below:
$F e_{2} O_{3} + C \to F e + C O_{2}$
How much $80 %$ pure iron could be produced from $120 k g$ of $90 %$ pure $F e_{2} O_{3}$ ?

94.5 k g

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60.48 k g

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116.66 k g

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120 k g

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Solution

The correct option is A $94.5 k g$
Balanced reaction is $2 F e_{2} O_{3} + 3 C \to 4 F e + 3 C O_{2}$ .
Number of moles of $F e_{2} O_{3}$ is $(\frac{120 \times 1000}{2 \times 56 + 48}) \times \frac{90}{100}$ .
Mass of $80 %$ pure iron produced is $\frac{120 \times 1000 \times 0.9}{2 \times 56 + 48} \times \frac{2 \times 56}{0.8}$ $= 94500 g$ or $94.5 k g$ .

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Similar questions

In the preparation of iron from haematite $(F e_{2} O_{3})$ by the reaction with carbon $F e_{2} O_{3} + C ⟶ F e + C O_{2}$
How much 80% pure iron could be produced from 120 kg of 90% pure $F e_{2} O_{3} ?$

In the extraction of Iron from Haematite Fe₂O₃, the ore is calcinated before ______ .

In the extraction of Iron from Haematite Fe₂O₃, the ore is calcinated before smelting – True or False?

Q. How many moles of iron can be made from the excess of

F e_{2} O_{3}

using 16 mol of carbon monoxide in the following reaction?

F e_{2} O_{3} + 3 C O \to 2 F e + 3 C O_{2}

Q. Iron (III) oxide can be reduced with

C O

to form metallic iron as described by unbalanced chemical reaction.

F e_{2} O_{3} + C O \to F e + C O_{2}

The number of moles of

C O

required to form one mole of

F e

from its oxide is:

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In the preparation of iron from haematite (Fe2O3) by the reaction with carbon as given below:Fe2O3+C→Fe+CO2 How much 80% pure iron could be produced from 120kg of 90% pure Fe2O3?

The correct option is A 94.5kgBalanced reaction is 2Fe2O3+3C→4Fe+3CO2.Number of moles of Fe2O3 is (120×10002×56+48)×90100.Mass of 80% pure iron produced is 120×1000×0.92×56+48×2×560.8=94500g or 94.5kg.

In the preparation of iron from haematite $(F e_{2} O_{3})$ by the reaction with carbon as given below:
$F e_{2} O_{3} + C \to F e + C O_{2}$
How much $80 %$ pure iron could be produced from $120 k g$ of $90 %$ pure $F e_{2} O_{3}$ ?