Question

One step in the conversion of aluminum ore into aluminum metal involves the synthesis of cryolite (Na3AlF6) in 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 1.00 kg of Na3AlF6?

Answer 1

Number of moles of Na3AlF6 = 1000 g / 209.94 g/mol = 4.76 mole

From the balanced equation we can say that

1 mole of Na3AlF6 requires 3 mole of NaAlO2 so

4.76 mole of Na3AlF6 will require

= 4.76 mole of Na3AlF6 *(3 mole of NaAlO2 / 1 mole of Na3AlF6)

= 14.3 mole of NaAlO2

number of moles of NaAlO2 = mass of NaAlO2 / molar mass of NaAlO2

14.3 mole = mass of NaAlO2 / 81.97 g/mol

mass of NaAlO2 = 1172 g

Therefore, the mass of NaAlO2 required would be 1172 g