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?

Answer 1

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

102g             240g             240 g                               420g  

                                         60.4 x10^3 g                        ?

here limiting reagent is HF the product based on HF

240g HF gives ------------------->420 g Cryolyte

60.4 x10^3 g HF gives = 60.4 x10^3 g x 420 / 240 = 105700 g

crylote mass = 105.7 kg

excess reactanta = no reagent is in excess

102 g Al2O3 -------------> 240 g HF

x g Al2O3 ---------------> 60.4 x10^3 g

x = 25.67 kg

total Al2O3 consumed.

total NaOH consumed .

excess reactant left = 0