Answer to Question #74779 in Inorganic Chemistry for Nicole Cuellar
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?
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.