Question

Determine the limiting reagent if 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH.

Al2(SO3)3 + 6 NaOH ------> 3 Na2SO3 + 2 Al(OH)3

Answer 1

a) moles Al2(SO3)3 = 10.0 g x 1 mol/294 g = 0.0340 moles Al2(SO3)3

moles NaOH = 10.0 g x 1 mol/40 g = 0.25 moles NaOH

Divide each by the coefficient in the balanced equation:

0.0340 Al2(SO3)3 /1 = 0.0340 moles

0.25 NaOH/6 = 0.0417 moles  

Al2(SO3)3 is limiting as it is in the least supply

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