Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.
<span> 2 Al(OH)</span>₃ → Al₂O₃ + 3 H₂O
According to equation at STP,
67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃
Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L
X =
75.44 g of Al(OH)₂Result: 75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry
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