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.

1.) Balance the equation

- AlO3(s)+NaOH(l)+HF(g)-->Na3AlF6+H2O(g)

2.)If 17.5 kilograms of Al2O3(s), 51.4 kilograms of NaOH(l), and 51.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

3.)Which reactants will be in excess, (Al2O3, NaOH, or HF)

4.)What is the total mass of the excess reactants left over after the reaction is complete in KG?

Explanation / Answer

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

(17.5 kg Al2O3) / (101.9614 g Al2O3/mol) x (2/1) = 0.343 kmol Na3AlF6
(51.4 kg NaOH) / (39.9971 g NaOH/mol) x (2/6) = 0.428 kmol Na3AlF6
(51.4 kg HF) / (20.0064 g HF/mol) x (2/12) = 0.428 kmol Na3AlF6

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

(0.343 kmol Na3AlF6) x (209.9413 g/mol) = 72 kg Na3AlF6 produced

(17.5 kg Al2O3) / (101.9614 g Al2O3/mol) x (6/1) x (39.9971 g NaOH/mol) =
41.2 kg NaOH reacted
(17.5 kg Al2O3) / (101.9614 g Al2O3/mol) x (12/1) x (20.0064 g HF/mol) =
41.2 kg HF reacted

(51.4 kg NaOH initially - 41.2 kg NaOH reacted) +
(51.4 kg HF initially - 41.2 kg HF reacted) = 20.4 kg total excess reactants left over

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.