Answer to Question #334416 in General Chemistry for Han

Question

What mass of Al(OH)3 is produced if 22.7g of NaOH is consumed?

Answer 1

Solution:

The molar mass of NaOH is 40 g/mol

Therefore,

Moles of NOH = [22.7 g NaOH] × [1 mol NaOH / 40 g NaOH] = 0.5675 mol NaOH

Balanced chemical equation:

Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃

According to stoichiometry:

6 mol of NaOH produce 2 mol of Al(OH)₃

Thus, 0.5675 mol of NaOH produce:

[0.5675 mol NaOH] × [2 mol Al(OH)₃ / 6 mol NaOH] = 0.1892 mol Al(OH)₃

The molar mass of Al(OH)₃ is 78 g/mol

Therefore,

Mass of Al(OH)₃ = [0.1892 mol Al(OH)₃] × [78 g Al(OH)₃ / 1 mol Al(OH)₃] = 14.76 g Al(OH)₃

Mass of Al(OH)₃ = 14.76 g = 14.8 g

Answer: 14.8 grams of Al(OH)₃ are produced

Short form solution:

Balanced chemical equation:

Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃

According to stoichiometry:

Mass of Al(OH)₃ = [22.7 g NaOH] × [1 mol NaOH / 40 g NaOH] × [2 mol Al(OH)₃ / 6 mol NaOH] × [78 g Al(OH)₃ / 1 mol Al(OH)₃] = 14.755 g Al(OH)₃ = 14.8 g Al(OH)₃