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.
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