Cryolite, Na AlF (s), an ore used in the production of aluminum, can be synthesized using...

The unbalanced reaction we can write is

There are two Al on left and one Al on right. Hence, we will multiply 2 on Na₃AlF₆ to balance the number of Al on both side.

Now, there are 6 Na on right and 1 Na on left. Hence, we multiply 6 on NaOH to balance the number of Na on both side.

Now, there are 9 O on left and 1 O on right. Hence, we multiply 9 on H₂O to balance the number of O atoms.

Now, there are 12 F on right and 1 F on left. Hence, we multiply 12 on HF to balance the number of F on both sides.

Hence, the balanced reaction to form cryolite is

Now, we will calculate the number of moles of each substance taken

Amount of Al₂O₃ taken = 15.6 kg = 15600 g

Molar mass of Al₂O₃ = 101.96 g/mol

Hence, the number of moles of Al₂O₃ taken is

Amount of NaOH taken = 55.4 kg = 55400 g

Molar mass of NaOH = 40.00 g/mol

Hence, number of moles of NaOH taken is

Amount of HF taken = 55.4 kg = 55400 g

Molar mass of HF = 20.00 g/mol

Hence, the number of moles of HF is

Each mole of Al₂O₃ reacts with 6 moles of NaOH. Hence, the number of moles of NaOH that will react with 153 mol of Al₂O₃ is

Similarly, each mole of Al₂O₃ reacts with 12 moles of HF. Hence, the number of moles of HF that will react with 153 mol of Al₂O₃ is

Since we have 1385 mol of NaOH and 2770 mol of HF, we have more than the amount required to react for NaOH and HF.

Hence, the limiting reactant is Al₂O₃ and excess reactants are HF and NaOH.

Now, 1 mol of Al₂O₃ forms 2 moles of cryolite. Hence, the number of moles of cryolite formed can be calculated as

Molar mass of Na₃AlF₆ = 209.94 g/mol

Hence, the mass of cryolite formed can be calculated as

Hence,

Also, the reactants that are in excess are

Now, total moles of NaOH left unreacted = Total moles of NaOH - total moles reacted = 1385 mol - 918 mol = 467 mol NaOH

Molar mass of NaOH = 40.00 g/mol

Hence, mass of NaOH left unreacted =

Total moles of HF left unreacted = Total moles of HF - total moles of HF reacted = 2770 mol - 1836 mol = 934 mol

Molar mass of HF = 20.00 g/mol

Hence, mass of HF left unreacted =

Hence, the total mass of of excess reactants left over after the reaction is complete is

Hence,

Note answers are rounded to three significant figures.

Al2O3(s) + NaOH + HFq) + Naz AlF6 + H2O(g)

Al2O3(s) + NaOH + HFq) + 2Naz AlF6 + H2O (9)

Al2O3(s) + 6NaOH + HF) + 2Naz AlF6 + H2O (9)

Al2O3(s) + 6NaOH + HFq) 2Naz AlF6 +9H20 (9

Al2O3(s) + 6NaOH) +12HF) + 2Naz AlF6 +9H20)

Al2O3(s) + 6NaOH) +12HF) + 2Naz AlF6 +9H20)

mass molar mass 15600 g 101.96 g/mol 153 mol

mass molar mass 55400 g 40.00 g/mol 1385 mol

55400 g mass molar mass 2770 mol 20.00 g/mol

6 mol NaOH x 153 mol Al2O3 = 918 mol NaOH 1 mol AlO3

12 mol HF - * 153 mol Al2O3 = 1836 mol HF 1 mol Al,03

2 mol Naz AlF6 2 x 153 mol Al2O3 = 306 mol Na3AlF 1 mol ALO3

209.94 g -- x 306 mol Na3AlF6 ~ 64241 g 1 mol Naz AlF61 64.2 kg

mass of cryolite formed = 64.2 kg

NaOH

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40.00 g/mol x 467 mol = 18680 g 18.68 kg

20.00 g/mol x 934 mol = 18680 g 18.68 kg

18.68 kg + 18.68 kg = 37.36 kg 37.4 kg

total mass of excess reactant = 37.4 kg