The volume of $H_{2}$ at STP required to completely reduce $160$ g of $F e_{2} O_{3}$ is:

3 \times 22.4

litres

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2 \times 22.4

litres

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22.4

litres

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11.2

litres

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Solution

The correct option is A $3 \times 22.4$ litres
$F e_{2} O_{3} + 3 H_{2} \to 2 F e + 3 H_{2} O$
The molar mass of $F e_{2} O_{3}$ is $2 (56) + 3 (16) = 160$ g/mol.
160 g of $F e_{2} O_{3}$ corresponds to $\frac{160 g}{160 g / m o l} = 1 m o l$
Complete reduction of 1 mole of $F e_{2} O_{3}$ will require 3 moles of $H_{2}$ .
3 moles of $H_{2}$ will occupy a volume of $3 \times 22.4$ litres at STP.

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The volume of H2 at STP required to completely reduce 160 g of Fe2O3 is:

The correct option is A 3×22.4 litresFe2O3+3H2→2Fe+3H2OThe molar mass of Fe2O3 is 2(56)+3(16)=160 g/mol.160 g of Fe2O3 corresponds to 160g160g/mol=1molComplete reduction of 1 mole of Fe2O3 will require 3 moles of H2.3 moles of H2 will occupy a volume of 3×22.4 litres at STP.

The volume of $H_{2}$ at STP required to completely reduce $160$ g of $F e_{2} O_{3}$ is: