A student will prepare a significant quantity of a gold compound which will be used in her research project, as well as other projects in her research group. The first step calls for the reaction of the gold metal with aqua regia (a 3:1 v/v mix of concentrated hydrochloric acid and concentrated nitric acid). In her procedure, she is to use 6 times as much aqua regia as the stoichiometric quantity required. A graduate student performed the calculation and indicated that aqua regia mixed according to directions has a nitrate concentration of 3.08 M and a chloride concentration of 6.96 M, while the hydrogen ion concentration is 10.04 M. The reaction is,
Au(s) + NO3-(aq) + Cl-(aq) + H+(aq) NO2(g) + AuCl4-(aq) + H2O(l)
The piece of gold wire, that is, the starting material, weighs 9.562 grams. Calculate the volume, in mL, of aqua regia which is required, based on the consideration stated above.
Au(s) + NO3-(aq) + Cl-(aq) + H+(aq) NO2(g) + AuCl4-(aq) + H2O(l)
The piece of gold wire, that is, the starting material, weighs 9.562 grams. Calculate the volume, in mL, of aqua regia which is required, based on the consideration stated above.
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Au(s) + 3 NO3-(aq) + 4 Cl-(aq) + 6 H+(aq)--------> 3 NO2(g) + AuCl4-(aq) + 3 H2O(l)
? mole Au = 9.562 g Au x 1 mole / 197.0 g/mole = 0.04854 mole
mole HNO3 = 0.04854 mol Au x 3 mol HNO3 / mol Au = 0.1456 mole HNO3
amt. used = 6 x 0.1456 mole HNO3 = 0.8737 mole HNO3
? L HNO3 = 0.8737 mole HNO3 x 1 L / 3.08 mole HNO3 = 0.284 L
? mole Au = 9.562 g Au x 1 mole / 197.0 g/mole = 0.04854 mole
mole HNO3 = 0.04854 mol Au x 3 mol HNO3 / mol Au = 0.1456 mole HNO3
amt. used = 6 x 0.1456 mole HNO3 = 0.8737 mole HNO3
? L HNO3 = 0.8737 mole HNO3 x 1 L / 3.08 mole HNO3 = 0.284 L