Question

Cryolite, Na, AIF (s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: ALO3(s) + NaOH(1) + HF (8) Na, A1F + H20(g) If 14.7 kg of AL,O,(s), 58.4 kg of NaOH(1), and 58.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced? mass of cryolite produced: kg Na, AIF, Which reactants will be in excess? NaOH ALO, OHF What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants:

Answer 1

Al₂O2 + 6Nooh + 12 HF – 2 Na, Alfo + 9H₂O Imole smole 12 mole No of moles a gomolewear mars! Given given mars 144.17indes 101.96 Alsuz = 14.7 kg = 14700g, moles Noon = 58.4kg - 58400 g moles, 58400 - 1460.10 moles 39-997 HF = 58.4 kg, 58400g, mole, S8400 2918.54 moles limiting ragent moles yaho Sticorometimes I his which have minimum is limiting reagent product depend on its concentation Noon HF A1203 14417 moles 2918.54 moles 1460.10 ouls 6 moto 6 miles I meus 243.2 243.35 144,17 these are eares It is limiting reagen Imole Alzo, produce a amole alazarte * 40.17 mole produce > 2x 144.17 2 286.34 moles

288.34 mole have = 288.34 X 209.94 NOAIE 260534.099 g мал өв случаи - fo.S3Fa Ng 04 - Both are in exces HF excess readent left over Imole Alo require 6 male Noon 144.17 mole require a 144.17X 6 - 865.02 moles the reacted 146010 - 865.02 a 595.08 mole It has 598.08 x 39.999 23785.34 223.789

I mole of Al₂O3 require 12 mole F jua mole teque a 12x14419 2 1730.04 mole unhealted - 2918.54-1730.04 (excees). 2 1188.5 mole mars 1188.5x20.01 2 2 3761.8 G 1 молоk uf 2 23. 18 in excey