Part B: Total mass of excess reactants If 16.8 kg of Al₂O3(s), 52.4 kg of NaOH(1), and 52.4 kg of HF(g) react completely, how many kilograms of cryolite willbe produced?mass of cryolite produced:Which reactants will be in excess?✓ NaOH✓HF☐A1₂0369.18What is the total mass of the excess reactants left over after the reaction is complete?total mass of excess reactants:68.63kg Na3AlF6kg Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.Balance the equation for the synthesis of cryolite.equation: Al₂O3(s) + 6NaOH(1) + 12HF(g)mass of cryolite produced: 69.18If 16.8 kg of Al₂O3(s), 52.4 kg of NaOH(1), and 52.4 kg of HF(g) react completely, how many kilograms of cryolite willbe produced?Which reactants will be in excess?✓ NaOH-✓ HF2Na3AlF6 + 9H₂O(g)kg Na3AlF6

The given balanced equation is- Al2O3 (s)+6NaOH (l) + 12HF (g)→2Na3AlF6 +9H2O (g)To find the mass of excess reagent, first convert the mass of reactants into their moles respectively. Mass of Al2O3 = 16.8 kg = 16800 g Molar mass of Al2O3 = 101.96 g/mol Thus, moles of Al2O3 are- Moles of Al2O3=MassMolar mass=16800 g101.96 g/molMoles of Al2O3=164.77 moles Mass of NaOH = 52.4 kg = 52400 g Molar mass of NaOH = 40.0 g/mol Thus, moles of NaOH are- Moles of NaOH=MassMolar mass=52400 g40.0 g/molMoles of NaOH=1310 moles Mass of HF = 52.4 kg = 52400 g Molar mass of HF = 20.01 g/mol Thus, moles of HF are- Moles of HF=MassMolar mass=52400 g20.01 g/molMoles of HF=2618.7 moles Now, as per the reaction, 1 mole of Al2O3 needs 6 moles of NaOH and 12 moles of HF. Thus, 164.77 moles of Al2O3 needs - Moles of NaOH = 6×164.77 moles =988.62 molesMoles of HF = 12×164.77 moles =1977.2 moles Thus, Al2O3 is the limiting reagent and NaOH and HF are in excess. Amount of NaOH left after reaction with 164.77 moles of Al2O3 = 1310 - 988.62 = 321.38 moles Amount of HF left after reaction with 164.77 moles of Al2O3 = 2618.7-1977.2 = 641.5 molesThus, Mass of excess NaOH = Moles of remaining NaOH ×Molar mass of NaOH Mass of excess NaOH =321.38 moles×40.0 g/mol =12855.2 g = 12.855 kg≃12.86 kg Mass of excess HF = Moles of remaining HF ×Molar mass of HF Mass of excess HF =641.5 moles×20.05 g/mol =12862.1 g = 12.862 kg≃12.86 kg Thus, Total mass of the excess reactants =12.86 kg + 12.86 kg = 25.72 kg Hence, total mass of the excess reactants left after the completion of reaction = 25.72 kg.