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Assignment Score: 233/2000 Resources Hint Check Answer Question 14 of 20 Cryolite, Na, AIF (S), an ore used in the production
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Answer #1

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

Assignment Score: Give Up? Hint 233/2000 Resources Check Answer < Question 14 of 20 Cryolite, Na AlF,(s), an ore used in the

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