Answer to Question #74779 in Inorganic Chemistry for Nicole Cuellar

Question #74779
If 11.4 kilograms of Al2O3(s), 51.4 kilograms of NaOH(l), and 51.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?
1
Expert's answer
2018-03-20T08:35:56-0400
Al2O3(s) + 6NaOH(l) + 12HF(g) → 2Na3AlF6(s) + 9H2O(g)

n(Al2O3) = m/M = 11400g / 102g/mol = 111.765 mol

n(NaOH) = m/M = 51400g / 40g/mol = 1285 mol

n(HF) = m/M = 51400g / 20g/mol = 2570 mol

Al2O3 is the limiting reactant. So NaOH and HF are in excess.

n(Na3AlF6) = 2*n(Al2O3) = 2*111.765 = 223.53 mol

M(Na3AlF6) = 23*3 + 27 +19*6 = 210 g/mol

m(Na3AlF6) = n*M = 223.53mol*210g/mol = 46941.3 g = 46.94 kg

46.94 kg of cryolite will be produced.

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