Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesi

Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation.

a)If 13.8 kilograms of Al2O3(s), 56.4 kilograms of NaOH(l), and 56.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

b)Which reactants will be in excess?

Question 8 of B Incorrect Incorrect 8 Map University Science Books General Chemistry 4th Edition Rock presented by Sapling Learning Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Tip: If you need to clear your Balance the equation. work and reset the equation, AI,O, (s)+6Nao H(I) 12HFlg) click the button that looks U like two red arrows. If 13.8 kilograms of Al203(s), 56.4 kilograms of NaoH (l, and 56.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced? Number kg Na, AIF. Which reactants will be in excess? at is the total mass of the excess reactants left over arter the reaction is complete? Al203 Number M NaOH kg M HF A Previous 3 Give Up & View Solution Check Answer Next Exit

Explanation / Answer

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

(14.6 kg Al2O3) / (101.9614 g Al2O3/mol) x (2/1) = 0.28638 kmol Na3AlF6
(56.4 kg NaOH) / (39.9971 g NaOH/mol) x (2/6) = 0.46541 kmol Na3AlF6
(56.4 kg HF) / (20.0064 g HF/mol) x (2/12) = 0.45105 kmol Na3AlF6

Since Al2O3 produces the least amount of product, Al2O3 is the limiting reactant. NaOH and HF are in excess.

(0.28638 kmol Na3AlF6) x (209.9413 g/mol) = 60.12kg Na3AlF6 produced

(14.6 kg Al2O3) / (101.9614 g Al2O3/mol) x (6/1) x (39.9971 g NaOH/mol) = 34.36 kg NaOH reacted
(14.6kg Al2O3) / (101.9614 g Al2O3/mol) x (12/1) x (20.0064 g HF/mol) = 34.37 kg HF reacted

(56.4 kg NaOH initially - 34.36 kg NaOH reacted) +
(56.4 kg HF initially - 34.37 kg HF reacted) = 44.07 kg

44.07 * 2=88.1 kg

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