Cryolite, Na_3 AlF_6(s), an ore used in the production of aluminum, can be synth

Question

Cryolite, Na_3 AlF_6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation. Al_2 O_3(s) + 6NaOH(l) + 12HF(g) rightarrow 2Na_3 AlF_6 + 9H_2 O(g) Tip: If you need to clear your work and reset the equation, click the button that looks like two red arrows. If 16.2 kilograms of Al_2 O_3(s), 59.4 kilograms of NaOH(l), and 59.4 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

Al2O3 + 6 NaOH + 12 HF 2 Na3AlF6 + 9 H2O

No.ofmoles of Na3AlF6= (16.2kg Al2O3) / (101.96137 g Al2O3/mol) x (2 mol Na3AlF6 / 1 mol Al2O3) =
0.32 kmol Na3AlF6
No.ofmoles of NaOH = (59.4 kg NaOH) / (39.99715 g NaOH/mol) x (2 mol Na3AlF6 / 6 mol NaOH) =
0.49 kmol Na3AlF6
No.ofmoles of HF= (59.4 kg HF) / (20.00635 g HF/mol) x (2 mol Na3AlF6 / 12 mol HF) =
0.48 kmol Na3AlF6

A.
(0.32 kmol Na3AlF6) x (209.94127 g Na3AlF6/mol) = 67.2 kg Na3AlF6 produced
B.
So Al2O3 is the limiting reactant, and NaOH and HF are in excess.

C.
Calculate the mass of water produced, then use the Law of Conservation of Mass to determine what's left over:

(16.2 kg Al2O3) / (101.96137 g Al2O3/mol) x (9 mol H2O / 1 mol Al2O3) x
(18.01532 g H2O/mol) = 25.9 kg H2O

16.2 kg Al2O3 + 59.4 kg NaOH + 59.4 kg HF - 67.2 kg Na3AlF6 - 25.9 kg H2O =
41.9kg of excess NaOH and HF

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