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: A1,0,(s) + NaOH(1) + HF(g) Na A1F + H,O(g) If 16.4 kg of A1,0,(s), 60.4 kg of NaOH(1), and 60.4 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? HF NaOH ALO What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants: See MRG - General Chen

Answer 1

das given amount. molan mads 0 1604 kg A203 ② 60.4 kg (3) 60.4kg solution Stehlį find moles of each compaund. a) males of Al2 O3 = mass = 16400g = 160.78mt 102 g/mol 6) males of Naon = 60400g = 1510 mol 40g/mal c) males of HF = 60400g = 3020 mol 20g/mol Step in find limiting Roegento @ mac balanced Reaction - Al2O3 +12 HF + 6 Naon - 2 Naz Alfo o moles per coff of Alz8z = meles = 160.78 mol=160078 ceffo © moles per cafz. of Naon = 1510 mol - 25/067 3 males peu Coff, of hf = 3020 mol = 251067 els we can see males per lafficient of M203 is lowest. so limiting reagent - Alz 03 12

step (111) find mass of Cryolite, Naon, HF that reacte En son. 0 applied PoAC Brele con balanced wreaction. cell₂O₃ + 12 HF + 6 Naon - 2 NO₂ Alfo males of Na Alto - moles of Al2O₃ officient Cofficient mass of NU Alfo - 160.78 mol molau mass X2 = 1 - 160.78 mass of Naz Alf6 209.9 4X2 67.51 kg/ mass of Na, Alfo = 2 again POAC for Naon . . mass of Naon - moles of Al2O3 molau mass (40) X Cofficient officient mass of Naon & 160.78 40X6 mass of Naona 38.59 kg

POAC for HF mass of HF Molave madt (20) x Cofficient - males of M₂ O3 Coffierent. = 160.78 mol mass of HF 20812 mass of HF = 38.59 kg Step Tu Mass - zmaning - find the mass of excess suggent. Remaning maxs of NaOH + Remahing massof HP Enitial maus - mass that react in Rony Maud Mass = Mars = (60.4-38.5g) + (6004-38.59) 21.81 + 21.81 Mars = 43.62 kg Remaning mass = 43.62 kg Thank you, May it helps you .