Question

How many moles of aluminum chloride and sodium hydroxide would be needed to produce 7.80 moles of aluminum hydroxide

Answer 1

The reaction between aluminium chloride and sodium hydroxide can be written as following:

AlCl₃ + 3NaOH = Al(OH)₃ + 3NaCl

As 7.80 moles of aluminium hydroxide are produced, 7.80 moles of aluminium chloride and 7.80 × 3 = 23.4 moles of sodium hydroxide are required.

Answer: 7.80 mol AlCl₃ and 23.4 mol NaOH.