1.Al2O3Al_2O_3Al2O3 was leached with alkali to get X. The solution of X on passing of gas Y, forms Z. X, Y and Z respectively area.X=Al(OH)3X = Al(OH)_3X=Al(OH)3, Y=CO2Y = CO_2Y=CO2, Z=Al2O3Z = Al_2O_3Z=Al2O3b.X=Na[Al(OH)4]X = Na[Al(OH)_4]X=Na[Al(OH)4], Y=SO2Y = SO_2Y=SO2, Z=Al2O3Z = Al_2O_3Z=Al2O3c.X=Al(OH)3X = Al(OH)_3X=Al(OH)3, Y=SO2Y = SO_2Y=SO2, Z=Al2O3⋅xH2OZ = Al_2O_3 \cdot xH_2OZ=Al2O3⋅xH2Od.X=Na[Al(OH)4]X = Na[Al(OH)_4]X=Na[Al(OH)4], Y=CO2Y = CO_2Y=CO2, Z=Al2O3⋅xH2OZ = Al_2O_3 \cdot xH_2OZ=Al2O3⋅xH2OLogin to continueOnly logged in users canattempt or see the solution.
