Question

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

Al2O3 (s) + 6NaOH (l) +12HF (g) --> 2Na3AlF6 + 9H2O (g) (I think I've already correctly done this)

A. if 10.5 kilograms of Al2O3(s), 52.4 kilograms of NaOH(l), and 52.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced

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

Answer 1

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

102 g          240g             144 g                                  420 g      

10.5kg         52.4kg            52.4kg                                 ?

HF is completely consumed means it is limiting reagent.

144 g HF----------------------> 420 g Na3AlF6

52.4 x 10^3 g HF--------------> 52.4 x 10^3 x 420 / 144 = 152833 g = 152.833 kg

152.833 kg cryolyte is produced

102 g Al2O3 --------------------> 114 g HF

x g Al2O3 --------------------> 52.4 x 10^3

x = 102 x 52.4 x 10^3 / 114

x = 46.7 kg of Al2O3 needed but we have 10.5 kg only so it is also reacted completely

240 g NaOH ---------------> 144 g HF

x g NaOH ---------------------> 52.4 x 10^3 g

x = 87.333 kg NaOH need so but have 52.4 kg so it is consumed in the reaction

B) no mass of excess reactnat left over