18 g of steam is passed over 2 g of iron-H2O - Fe - H2 - Fe2O3- is the excessive reagent- -Fe - 56 g-

First, balance the given reaction. 3H_2O+2Fe → 3H_2+Fe_2O_3 ∴ 3 mol nbsp;of H_2O require nbsp;2 mol nbsp;of Fe The ratio of moles of water to iron is 3/2. ...

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Byju's Answer

Standard XI

Chemistry

Limiting Reagent or Reactant

18 g of steam...

Question

$18 g$ of steam is passed over $2 g$ of iron. $H_{2} O + F e \to H_{2} + F e_{2} O_{3}$ . is the excessive reagent? [Fe = 56 g]

H_{2} O

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H_{2}

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F e

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F e_{2} O_{3}

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Solution

The correct option is A $H_{2} O$
First, balance the given reaction.
$3 H_{2} O + 2 F e \to 3 H_{2} + F e_{2} O_{3}$
$∴$ 3 mol of $H_{2} O$ require 2 mol of $F e$
The ratio of moles of water to iron is $\frac{3}{2}$ . If the actual ratio is lower than this, that means that there is a lower amount of water, making iron the excess reagent. If this is higher than 1.5, then water is the excess reagent. Let's calculate!
Moles of water = $\frac{18 g}{36 g / m o l} = 0.5 m o l$
Moles of iron = $\frac{2 g}{56 g / m o l} = 0.036 m o l$
Ratio - $\frac{0.5}{0.036} = 13.88$
The ratio is much higher than 1.5, making water the excess reagent.

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18g of steam is passed over 2g of Iron. $H_{2} O + F e \to H_{2} + F e_{2} O_{3}$ Which is the excessive reagent? [Fe = 55.85g]

18 g

of steam is passed over

2 g

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

. is the excessive reagent? [Fe = 56 g]

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(Atomic mass of Fe=56 g.

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{Fe}_{2} O_{3}

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18 g of steam is passed over 2 g of iron.H2O + Fe → H2 + Fe2O3. is the excessive reagent? [Fe = 56 g]

$18 g$ of steam is passed over $2 g$ of iron. $H_{2} O + F e \to H_{2} + F e_{2} O_{3}$ . is the excessive reagent? [Fe = 56 g]