Question

Cryolite, Na AlF (s), an ore used in the production of aluminum, can be synthesized using aluminum oxide Balance the equation for the synthesis of cryolite equation: Al03 (s) + NaOH(I) + HF(g) » Na3 AIF6H2O(g) If 15.6 kg of Al, 0,(s), 55.4 kg of N2OH(l), and 55.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced? kg Na3 AIF6 mass of cryolite produced: Which reactants will be in excess? NaOH Al203 HF What is the total mass of the excess reactants left over after the reaction is complete? total mass of excess reactants kg

Answer 1

The unbalanced reaction we can write is

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

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.

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

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.

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

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.

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

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.

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

Hence, the balanced reaction to form cryolite is

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

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

mass molar mass 15600 g 101.96 g/mol 153 mol

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

mass molar mass 55400 g 40.00 g/mol 1385 mol

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

55400 g mass molar mass 2770 mol 20.00 g/mol

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

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

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

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

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

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

Molar mass of Na₃AlF₆ = 209.94 g/mol

Hence, the mass of cryolite formed can be calculated as

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

Hence,

mass of cryolite formed = 64.2 kg

Also, the reactants that are in excess are

NaOH

${\color{Red} HF}$

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 =

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

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 =

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

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

18.68 kg + 18.68 kg = 37.36 kg 37.4 kg

Hence,

total mass of excess reactant = 37.4 kg

Note answers are rounded to three significant figures.