Need help for this question a, b, c portion 1. Aluminum sulfite reacts with sodium hydroxide to form sodium sulfite and aluminum hydroxide according to the followingequation.Al=27, S=32, H=1, O=16Al2(SO3)3 + 6 NaOH ------>3 Na2SO3 + 2 Al(OH)3a. Determine the limiting reactant if 20.0 g of Al2(SO3}3 reacts with 20.0 g of NaOH.b. Determine the number of moles and mass of Al(OH)3c. Determine the number of moles and mass of Na2SO3

The balanced chemical reaction is, Al2(SO3)3 + 6NaOH ------->3Na2SO3 + 2Al(OH)3 So, 1 moles of Al2(SO3)3 react with 6 moles of NaOH to produce 3 moles of Solute and 2 moles of Al(OH)3. Number of moles = mass/molar mass. So the number of moles of 20.0g of Al2(SO3)3 is, n = 20.0/294.15 n = 0.068 moles Similarly, the number of moles of 20.0 g NaOH is, n = 20.0/40.0 n = 0.500 moles a) according to the balanced chemical reaction, 1 moles of Al2(SO3)3 react with 6 moles of NaOH. So, 0.068 moles of Al2(SO3)3 require 6×0.068 = 0.408 moles of NaOH. But we have 0.500 moles of NaOH. Hence NaOH is the excess reagent and Al2(SO3)3 is the limiting reagent. So, limiting reagent is Al2(SO3)3 b) According to the balanced chemical reaction, 1 moles of Al2(SO3)3 react with 6 moles of NaOH to produce 3 moles of Solute and 2 moles of Al(OH)3. So, 0.068 moles of Al2(SO3)3 will produce 2×0.068 = 0.136 moles of Al(OH)3 Mass = number of moles × molar mass Mass = 0.136 × 78 Mass = 10.61 g So the answers are 0.136 moles of Al(OH)3 and 10.61 gc) According to the balanced chemical reaction, So, 1 moles of Al2(SO3)3 react with 6 moles of NaOH to produce 3 moles of Na2SO3. So, 0.068 moles of Al2(SO3)3 will produce 3×0.068 = 0.204 moles of Na2SO3 Mass = number of moles × molar mass Mass = 0.204 × 126.04 Mass = 25.71 g So the answers are 0.204 moles of Na2SO3 and 25.71 g