Al2O3(s) + 6 NaOH(l) +Â Â 12 HF(g)
-----------------> 2Na3AlF6 + 9 H2O(g)
102g           Â
240g           Â
240
g                             Â
420g Â
                                       Â
60.4 x10^3
g                      Â
?
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
