Final answer:
The weight of 2.04 moles of aluminum hydroxide (Al(OH)₃) is approximately 159.32 g.
Step-by-step explanation:
To find the weight of 2.04 moles of aluminum hydroxide (Al(OH)₃), we need to calculate the molar mass of Al(OH)₃ and then multiply it by the number of moles.
The molar mass of Al(OH)₃ can be calculated by adding up the atomic masses of each element in the compound:
Using these values, we calculate the molar mass of Al(OH)₃ as follows:
(26.98 g/mol) + 3(16.00 g/mol) + 3(1.01 g/mol) = 78.02 g/mol
Finally, we multiply the molar mass by the number of moles:
78.02 g/mol x 2.04 mol = 159.32 g
Therefore, the weight of 2.04 moles of aluminum hydroxide (Al(OH)₃) is approximately 159.32 g.