Answer to Question #311848 in Chemistry for grey tyrant

Question

4. NaOH react completely with Al(NO3)3.

c. What mass of aluminum hydroxide is produce from 25 ml 1.0 M Al(NO3)3.

Answer 1

Solution:

Moles of Al(NO₃)₃ = Molarity of Al(NO₃)₃ × Volume of Al(NO₃)₃

Moles of Al(NO₃)₃ = (1.0 M) × (0.025 L) = 0.025 mol

Moles of Al(NO₃)₃ = 0.025 mol

Balanced chemical equation:

3NaOH + Al(NO₃)₃ ⟶ Al(OH)₃ + 3NaNO₃

According to stoichiometry:

1 mol of Al(NO₃)₃ produces 1 mol of Al(OH)₃

Thus, 0.025 mol of Al(NO₃)₃ produces 0.025 mol of Al(OH)₃

The molar mass of Al(OH)₃ is 78 g/mol

Therefore,

Mass of Al(OH)₃ = [0.025 mol Al(OH)₃] × [78 g Al(OH)₃ / 1 mol Al(OH)₃] = 1.95 g Al(OH)₃

Mass of Al(OH)₃ = 1.95 g

Answer: 1.95 grams of aluminum hydroxide is produced