We can see from the equation that 2 moles of aluminum hydroxide produce 3 moles of water. We can also see that the molar mass of aluminum hydroxide is 78.00 g/mol.
We are given that 2.4 x 10^25 molecules of water are produced. We can convert this to moles by dividing the number of molecules by Avogadro's number:
2.4 x 10^25 molecules H2O / 6.022 x 10^23 molecules/mol H2O = 4.0 x 10^2 moles H2O
We can then use the molar ratio of aluminum hydroxide to water to calculate the mass of aluminum hydroxide that is needed:
4.0 x 10^2 moles H2O / 3 moles H2O/ 2 moles Al(OH)3 = 4.0 x 10^2 moles Al(OH)3
We can then use the molar mass of aluminum hydroxide to calculate the mass of aluminum hydroxide that is needed:
4.0 x 10^2 moles Al(OH)3 * 78.00 g/mol Al(OH)3 = 3.12 x 10^3 g Al(OH)3
Therefore, 3.12 x 10^3 g of aluminum hydroxide are needed to decompose in order to produce 2.4 x 10^25 molecules of water at STP.
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