Slide to the correct value of x to balance the given equation.

${Fe}_{2} O_{3} (s) + 3 H_{2} (g) \to 2 Fe (s) + {xH}_{2} O (l)$
3

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Solution

The correct option is A 3
A balanced chemical equation has an equal number of atoms on both the sides of a chemical reaction.

For the given equation: ${Fe}_{2} O_{3} (s) + 3 H_{2} (g) \to 2 Fe (s) + {xH}_{2} O (l)$
Reactants Products

Elements No. of atoms Elements No. of atoms

Fe 2 Fe 2

O 3 O x

H 6 H 2x

Here, Fe atoms are balanced on both sides. To balance H and O atoms, the value of $x$ should be 3.

Hence, the balanced equation is ${Fe}_{2} O_{3} (s) + 3 H_{2} (g) \to 2 Fe(s) + 3 H_{2} O(l)$

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Similar questions

{Fe}_{2} O_{3} (s) + 3 H_{2} (g) \to 2 Fe (s) + {xH}_{2} O (l)

x

M g_{3} N_{2} (a q) + x H_{2} O (l) \to 3 M g (O H)_{2} (a q) + 2 N H_{3} (g)

Fe2O3+CO gives 2Fe+CO2

BALANCE THE EQUATION

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Δ H

F e_{2} O_{3 (s)} + 3 H_{2 (g)} \to 2 F e_{(s)} + 3 H_{2} O_{(l)}

Δ H_{300} = - 33.29 k J m o l^{- 1}

C_{p}

F e_{2} O_{3 (s)}, F e_{(s)}, H_{2} O (l)

H_{2 (g)}

103.8, 25.1, 75.3

28.8

J K^{- 1} m o l^{- 1}

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