Balance the following reaction,Al2O3⋅2H2O+xNaOH+H2O⟶xNaAl(OH)4What is the value of x?

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Balance the following reaction,

${Al}_{2} O_{3} \cdot 2 H_{2} O + x NaOH + H_{2} O ⟶ x Na [Al (OH)_{4}]$

What is the value of x?

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answer is 2.

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Detailed Solution

The reaction is,

${Al}_{2} O_{3} \cdot 2 H_{2} O + 2 NaOH + H_{2} O ⟶ 2 Na [Al ({OH}_{4}]$

Al must first be balanced on both sides.

${Al}_{2} O_{3} + NaOH + H_{2} O ⟶ 2 NaAl (OH)_{4}$

Na must now be balanced on both sides. ${Al}_{2} O_{3} + 2 NaOH + H_{2} O \to 2 NaAl (OH)_{4}$

H and O are now being balanced on both sides.

${Al}_{2} O_{3} + 2 NaOH + 3 H_{2} O ⟶ 2 NaAl (OH)_{4}$

In order to balance the equation, 2 moles of $NaOH$ are needed.

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