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$A l_{(s)} + F e_{2} O_{3 (s)} \to A l_{2} O_{3 (s)} + F e_{(s)}$
(atomic masses of $A l = 27 U, F e = 56 U, a n d O = 16 U$ )

$2 A l_{(s)} + F e_{2} O_{3 (s)} \to A l_{2} O_{3 (s)} + 2 F e_{(s)}$ , is a balanced equation.

Suppose that your are asked to calculate the amount of aluminium, required to get 1120 kg of iron by the above reaction.

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540 Kg

280 Kg

1120 Kg

270 Kg

answer is A.

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Detailed Solution

2 A l_{(s)} + F e_{2} O_{3 (s)} \to A l_{2} O_{3 (s)} + 2 F e_{(s)}

As per the balanced equation

$Aluminium \to I r o n$

$54 g \to 112 g$

$x ? \to (1120 \times 1000) g$

$∴ x g = \frac{(1120 \times 1000) g \times 54 g}{112 g}$

$= 10000 \times 54 g$

$= 540000 g o r 540 k g$

$∴$ To get 1120 kg of iron we have to use 540 kg of aluminium.

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