Calculate the mass, in grams, of Al(OH)3 required for this reaction:

Reactant(s) Products
Al(OH)3 ⟶ Al2O3 + H2O
mass: 21.8 grams 9.7 grams
Enter only a number for your answer.

Mass = 28.08

Explanation:

Given data:

Mass of Al₂O₃ = 21.8 g

Mass of water = 9.7 g

Mass of Al(OH)₃ = ?

Solution:

Chemical equation:

2Al(OH)₃   →   Al₂O₃ + 3H₂O

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 9.7 g/ 18 g/mol

Number of  moles = 0.54 mol

Number of moles of Al₂O₃:

Number of moles = mass/molar mass

Number of moles = 21.8 g/ 101.96 g/mol

Number of  moles = 0.21 mol

Now we will compare the moles of Al(OH)₃ with  Al₂O₃ and H₂O.

                          Al₂O₃       :          Al(OH)₃

                             1            :              2

                         0.21          :         2×0.21 = 0.42 mol

                         H₂O          :            Al(OH)₃

                            3            ;             2

                          0.54        :        2/3×0.54 = 0.36 mol

Mass of Al(OH)₃:

Mass = number of moles × molar mass

Mass = 0.36 mol × 78 g/mol

Mass = 28.08 g

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