Question

The manufacture of aluminum includes the production of cryolite (Na₃AlF₆) from the following reaction.

6 HF(g) + 3 NaAlO₂(s) → Na₃AlF₆(s) + 3 H₂O(l) + Al₂O₃(s)

How much NaAlO₂ (sodium aluminate) is required to produce 1.82 kg of Na₃AlF₆?

Answer 1

molar mass of Na3AlF6 = 209.9 g/mol
given mass = 1.82 kg
= 1820 g
number of mole of Na3AlF6 = (given mass)/(molar mass)
= 1820/209.9
= 8.67 mole

according to reaction
1 mole of Na3AlF6 is produced by 3 mole of NaAlO2
8.67 mole of Na3AlF6 is produced by (3*8.67) mole of NaAlO2
number of mole of NaAlO2 = (3*8.67)
= 26 mole

molar mass of NaAlO2 = 82 g/mol
mass of NaAlO2 required = (number of mole of NaAlO2)*(molar mass)
= 26*82
= 2132 g
= 2.132 kg
Answer : 2.132 kg