TS EAMCET · Chemistry · Some Basic Concepts of Chemistry

The amount of iron (Fe) in $g$ which can be produced from $600 \mathrm{~g}$ of magnetite ore is [Atomic mass of Fe : 55.8]

A 450
B 379
C 434
D 210

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Answer & Solution

Correct Answer

(C) 434

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Detailed explanation

$\mathrm{Fe}_3 \mathrm{O}_4+2 \mathrm{C} \longrightarrow 3 \mathrm{Fe}+2 \mathrm{CO}_2$ 1 mole of $\mathrm{Fe}_3 \mathrm{O}_4$ produce 3 moles of $\mathrm{Fe}$ so, $600 \mathrm{~g}$ of magnitite have $=\frac{600}{232}$ moles $=2.5 \mathrm{~mol}$. So, 2.5 moles of…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following

Column-1 Reaction		Column-2 Main product
A)	$4 {FeCl}_{3} + 3 K_{4} [Fe (CN)_{6}] \to$	I)	${Fe}_{4} {[Fe (CN)_{6}]}_{3}$
B)	${ZnCl}_{2} + 4 NaOH \to$	II)	${FeCl}_{2}$
C)	$2 {FeCl}_{3} + H_{2} S \to$	III)	$Zn (OH) Cl$
		IV)	${Fe}_{3} {[Fe (CN)_{6}]}_{3}$
		V)	${Na}_{2} {ZnO}_{2}$
		VI)	$FeS$

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The amount of iron (Fe) in \(g\) which can be produced from \(600 \mathrm{~g}\) of magnetite ore is [Atomic mass of Fe : 55.8]

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