Question

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

Al2O3+NaOH+HF-->Na3+AlF6+H20

If 16.3 kilograms of Al2O3(s), 59.4 kilograms of NaOH(l), and 59.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

What is the total mass of the excess reactants left over after the reaction is complete?

Answer 1

Given Reaction : Al₂O₃ + NaOH + HF --> Na₃AlF₆ + H₂O
First We can balance aluminum to give
Al₂O₃ + NaOH + HF --> 2 Na₃AlF₆ + H₂O
Then we can balance sodium to give
Al₂O₃ + 6 NaOH + HF --> 2 Na₃AlF₆ + H₂O
Then Fluorine
Al₂O₃ + 6 NaOH + 12 HF --> 2 Na3AlF₆ + H₂O
Then the water, there are 9 oxygens and 18 hydrogens on the left which make 9 waters on the right so the final balanced equation is
Al₂O₃ + 6 NaOH + 12 HF --> 2 Na₃AlF₆ + 9 H₂O

Note. Water and oxygen are most common and easy to balance in a raection. So finally we can consider them while balancing the equation.

Mole and weight calculation is given in the following picture

Mass of excess Reagent:

Excess of NaOH= Total mol- reacted

= 1485-958.8

= 526.2 mol

Therefore Weight of NaOH = 526.2 x 40g (M.Wt)

= 21048g

   =21.048 Kg

Excess Of HF = Total mol- reacted

= 2970 - 1917.6

= 1052.4 mol

Therefore Weight of HF  = 1052.4 x 20 (M.Wt)

= 2104.8 g

= 2.105 Kg