Map General Chemistry 4th Editionp . Gallo University Science Books presented by
ID: 553360 • Letter: M
Question
Map General Chemistry 4th Editionp . Gallo University Science Books presented by Sapling Learning Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide ip: If you need to clear your work and reset the equation, click the button that looks like two red arrows Balance the equation ALO,(s) + NaOH(1) + HF(g) Na,AlF,+H2O(g) If 17.0 kilograms of Al203(s), 57.4 kilograms of NaOH(), and 57.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced? Number kg NazAIF Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete? Al203 NaOH HF NumberExplanation / Answer
mass of Al2O3 = 17kg molecular weight of Al2O3 = 102g/mol
mass of NaOH = 57.4Kg molecular weight of NaOH = 40g/mol
mass of HF = 57.4Kg molecular weight of HF = 20g/mol
mole = weight given/molecular weight
moles of Al2O3 = 17000/102
moles of Al2O3 = 166.67
moles of NaOH = 57400/40
moles of NaOH = 1435
moles of HF = 57400/20
moles of HF = 2870
Al2O3 + 6 NaOH + 12HF---------------------> 2Na3AlF6 + 9 H2O
Al2O3 is limiting reagent in the given reaction,
so, 1 mole of Al2O3 produce 2mole of Na3AlF6
so 166.67 mole of Al2O3 will produce (166.67*2)mole of Na3AlF6
mass of Na3AlF6 produced = moles of Na3AlF6 * molecular mass of Na3AlF6
mass of Na3AlF6 produced = 166.67*210*2
mass of Na3AlF6 produced = 70001.4g (70.0014kg)
1 mole of Al2O3 produces 9 mole of H2O so, 166.67 mole will produce 1500.03 mole of H2O
mass of H2O formed = moles of H2O* molecular mass of H2O
mass of H2O formed = 1500.03*18
mass of H2O formed = 27000.54g (27.00054Kg)
total mass left over in excess after complete reaction = mass of reactant - mass of product
total mass left over in excess after complete reaction = (131.8 - 97.00194)kg
total mass left over in excess after complete reaction = 34.79806Kg