Question

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

Al2O3(s) + NaOH(l) + HF(g) ---> Na3AlF6 + H2O(g)

If 11.3 kilograms of Al2O3(s), 60.4 kilograms of NaOH(l), and 60.4 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 in kilograms?

Answer 1

The balanced chemical equation is

The molar masses of Al2O3, NaOH and HF are 102 g/mol, 40 g/mol and 20 g/mol respectively.
11.3 kilograms of Al2O3(s) corresponds to kmol
60.4 kilograms of NaOH(l) corresponds to kmol
and 60.4 kilograms of HF(g) corresponds to kmol.
1 kmol of Al2O3 will react with 6 kmol NaOH and 12 kmol HF.
0.1108 kmol of Al2O3 will react with 0.665 kmol NaOH and 1.33 kmol HF
Thus Al2O3 is the limiting reactant and HF and NaOH are the excess reactants.
Number of kilo moles of cryolite obtained kmol
The molar mass of cryolite is 210 g/mol.
Mass of cryolite obtained kg
Kmoles of NaOH remainining kmol
Mass of NaOH remaining kg
Kmoles of HF remaining kmol
Mass of HF remaining kg
Total mass of excess reactants remaining kg.