Question

Al2O3(s) + 6NaOH(l) + 12HF(g) = 2Na3AlF6 + 9H2O(g)

If 10.9 kilograms of Al2O3(s), 50.4 kilograms of NaOH(l), and 50.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

A2O3 moles = mass of Al2O3 in grams / Molar mass of AL2O3 = ( 10.9 x 1000) / 101.96 = 106.9

NaOH moles = ( 50.4 x 1000) / 39.997) = 1260.095

HF moles = ( 50.4 x 1000) / 20.01 = 2518.74

as per Al2O3 based on coeffeients in reaction we need NaoH moles = 6 x AL2O3 = 6 x 106.9 = 641.4

HF moles needed = 12 x 106.9 = 1282.8

but we see we have excess of HF and excess NaOH ,

hence Al2O3 is limiting reagent ( relatively less in moles compared to other)

now product Na3AlF6 moles = 2 x Al2O3 moles = 2 x106.9 = 213.8

Na3AlF6 mass = moles x molar mass = 213.8 x 209.94 = 44885 g = 44.885 kg = 44.9 kg