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 dа ta:

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