Balanced equation is:
Al2O3(s) + 6 NaOH (l) + 12 HF (g) → 2 Na3AlF6(s) + 9 H2O (g)
moles Al2O3 = 16400 g / 101.96128 g/mol = 160.84537189 mol
moles NaOH = 59400 g / 39.99711 g/mol = 1485.107299 mol
moles HF = 59400 g / 20.006343 g/mol = 2969.05836314 mol
Al2O3 is limiting reactant here
moles Na3AlF6 = 2 x 160.84537189 mol = 321.69074378 mol
mass of cryolite produced = 321.69074378 mol x 209.9412656 = 67.5361618811 kg ~ 67.5 kg
excess reactants are = NaOH & HF
Excess moles of NaOH = 1485.107299 - ( 6 x 160.84537189 ) = 520.03506766 mol
mass of excess NaOH = 520.03506766 mol x 39.99711 g/mol = 20.8 kg
Excess moles of HF = 2969.05836314 - ( 12 x 160.84537189 ) = 1038.91390046 mol
mass of excess HF = 1038.91390046 mol x 20.006343 g/mol = 20.8 kg
Total mass of excess reactant left over = 20.8 + 20.8 = 41.6 kg
Hope this helped you!
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