Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation. If 15.1 kilograms of Al2O3(s), 58.4 kilograms of NaOH(l), and 58.4 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?

Answer 1

Balanced equation -

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

If 15.1 kilograms of Al2O3(s), 58.4 kilograms of NaOH(l), and 58.4 kilograms of HF(g) react completely.

Moles of Al2O3 = 15100 g / 101.96 g/mol

= 148.1 moles

Moles of NaOH = 58400 g / 39.997 g/mol

= 1460.11 moles

Moles of HF = 58400 g / 20.01 g/mol

= 2918.54 moles

SInce, Al2O3 is the limiting reagent.

Moles of cryolite produced = 2 * 148.1

= 296.2 moles

Mass of cryolite produced = 296.2 moles * 209.94 g/mol

= 62184.228 g

= 62.184 kg

Moles of NaOH consumed = 148.1 * 6 = 888.6

Moles of HF consumed = 148.1 * 12 = 1777.2

HF(g) will be in excess.

Moles of HF in excess = 2918.54 - 1777.2

= 1141.34 moles

Mass of HF left over after the reaction is complete = 1141.34 moles * 20.01 g/mol

= 22838.21 g

= 22.83 kg