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 11.4 kilograms of Al2O3(s), 55.4 kilograms of NaOH(l), and 55.4 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

Balanced equation:

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

11.4 kilograms of Al2O3 = 11400 /101.96 = 111.807 Moles

55.4 kilograms of NaOH = 55400 /40 = 1385.1 Moles

55.4 kilograms of HF = 55400 / 20 = 2769.12 Moles

According to the above equation Al2O3 only limiting reagent. Hence it will produce (111.807 x 2) 223.61 Moles of cryolite.

223.61 moles of cryolite = 223.61 x 209.94 = 46945.86 gm or 46.945 kg

46.945 kg of cryolite will be produced

Excess reagent = NaOH and HF will be excess

670.84 Moles of NaOH and 1341.68 Moles of HF will be used in the reaction.

670.84 Moles of NaOH =   670.84 x 40 = 26.83kg

341.68 Moles of HF = 341.68 x 20 = 26.84 kg

Total weight of Excess reagent = 55.4 + 55.4 - 26.83 -26.84 = 57.13 kg

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