The manufacture of aluminum includes the production of cryolite (na3alf6) from the following reaction: 6hf+3naalo2=na3alf6+3h2o+al2o3 how much naalo2 (sodium aluminate) is required to produce 3.47 kg of na3alf6?

6HF  +  3  NaAlO2  =  Na3AlF6  +3H2O  +Al203
calculate  moles  of  Na3AlF6   formed

moles  =mass/molar  mass
molar  mass  of  Na3AlF6  =209.94  g/mol
convert 3.47Kg  to  grams  ,  that  is  3.47  x1000=  3470  grams
moles  is  therefore= 3470  g/ 209.94  g/mol  =  16.53  moles

by  use  of  mole  ratio  between  NaAlO2  to  Na3AlF6  which  is    3:1  therefore  the   moles  of  NaAlO2   is  (16.53  x3)/ 1=  49.59  moles
mass  of  NaAlO2 is  =moles  of  NaAlO2  x  molar  mass  NaAlO2

that   is  49.59  mole  x  81.97 g/mol =  4064.89 grams  or  4.065  Kg

I’m pretty sure it’s false i’m not sure