A chemist wants to extract the gold from 35.93 g of AuCl3 · 2 H2O (gold(III) chloride dihydrate) by electrolysis of an aqueous solution. What mass of gold could be obtained from this sample?

Question

A chemist wants to extract the gold from 35.93 g of AuCl3 · 2 H2O (gold(III) chloride dihydrate) by electrolysis of an aqueous solution. What mass of gold could be obtained from this sample? Answer in units of g

Answer 1

molar mass AuCl3 · 2 H2O = 339.4 g / mol

moles of AuCl3 · 2 H2O used = 35.93 g x 1 mol / 339.4 g = 0.10586 mols

moles Au present = 0.10586 mols x 1 mol Au / mol AuCl3 · 2 H2O = 0.10586 mols Au

atomic mass Au = 196.97 g / mol

mass Au = 0.10586 mols Au x 196.97 g / mol = 20.85 g Au (4 sig. figs.) ... assuming 100% extraction