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Solution: Aluminum sulfite reacts with sodium hydroxide to form sodium sulfite and aluminum hydroxide according to the following equation.Al2(SO3)3 + 6 NaOH → 3 Na 2SO3 + 2 Al(OH)3a) Determine the limiting reactant if 20.0 g of Al 2(SO3)3 reacts with 20.0 g of NaOH.     b) Determine the number of moles and mass of Al(OH) 3 produced. 

Problem

Aluminum sulfite reacts with sodium hydroxide to form sodium sulfite and aluminum hydroxide according to the following equation.

Al2(SO3)3 + 6 NaOH → 3 Na 2SO3 + 2 Al(OH)3

a) Determine the limiting reactant if 20.0 g of Al 2(SO3)3 reacts with 20.0 g of NaOH.

 

 

 

 

 

b) Determine the number of moles and mass of Al(OH) 3 produced.