Cryolite, Na_3AIF_6(s), an ore used in the production of aluminum, can be synthe

Question

Cryolite, Na_3AIF_6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation. Tip: If you need to clear work and reset button that 100 like two red arrows. Al_2O_3(s) + NaOH(l) + HF(g) rightarrow Na_3AIF-6 + H_2O(g) If 14.0 kilograms of Al_2O_3(s), 54.4 kilograms of NaOH(1), and 54.4 kilograms of HF (g) react completely, how many kilograms of cryolite will be produced? Which reactants will be in excess? What is the total mass of the excess reactants left over after the reaction is complete?

Explanation / Answer

Given the mass of Al2O3 = 14.0 kg = 14000 g

Molecular mass of Al2O3 = 101.96 g/mol

Hence moles of Al2O3 taken = mass / molar mass = 14000 g / 101.96 g/mol = 137.31 mol

Given the mass of NaOH = 54.4 kg = 54400 g

Molecular mass of NaOH = 40.0 g/mol

Hence moles of NaOH taken = mass / molar mass = 54400 g / 40.0 g/mol = 1360 mol

Given the mass of HF = 54.4 kg = 54400 g

Molecular mass of HF = 20.0 g/mol

Hence moles of HF taken = mass / molar mass = 54400 g / 20.0 g/mol = 2720 mol

The balanced chemical reaction is

Al2O3(s) + 6 NaOH(l) + 12 HF(g) --------- > 2Na3AlF6(s) + 9 H2O(g) (answer)

1 mol, ------ 6 mol, -------- 12 mol -------------- 2 mol ------------ 9 mol

1 mole of Al2O3(s) reacts with 6 mol of NaOH

Hence 137.31 mol Al2O3(s) that will react with the moles of NaOH

= 137.31 mol Al2O3(s) x ( 6 mol NaOH / 1 mol Al2O3)

= 823.86 mol NaOH (less than the initial mol of NaOH)

Since NaOH is remained after the cmplete reaction of Al2O3, Al2O3 is the limiting reactant and is exhausted comletely.

The amount of HF that will react completely with Al2O3

= 137.31 mol Al2O3(s) x (12 mol HF / 1 mol Al2O3)

= 1647.72 mol HF (less than the initial mol of HF taken)

Hence both NaOH and HF are in excess (answer)

moles of excess NaOH remained = 1360 mol - 823.86 mol = 536.14 mol

Hence mass of excess NaOH remained = 536.14 mol x 40.0 g/mol = 21446 g = 21.446 Kg

moles of excess HF remained = 2720 mol - 1647.72 mol = 1072.28 mol

Hence mass of excess NaOH remained = 1072.28 mol x 20.0 g/mol = 21446 g = 21.446 Kg

Hnece mass of excess reactant left over = 21.446 Kg + 21.446 Kg = 42.89 Kg (answer)

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