Gold, Au, is dissolved from rock by treating the rock NaCN in the presence oxyge

Question

Gold, Au, is dissolved from rock by treating the rock NaCN in the presence oxygen 4 Au(s) + 8 NaCN(aq) + O_2 (g)+ 2 H_2O (l) rightarrow 4 NaAu(CN)_2 (aq) + 4NaOH(aq) If you begin the reaction with 5.00 grams of O_2and 5.00 grams of NaCN (with unlimited amount of Au and water). What is the limited reagent and the Theoretical yield for NaAu(CN)_2(Molar mass = 272g/mol) 1.20 grams of NaAu(CN)_2 was recovered experimentally, what would be the %yield? Calculate the amount of excess reactant left over.

Explanation / Answer

a) moles of O2 = 5 / 32 = 0.156

moles of NaCN = 5 / 49 = 0.102

So, limited reagent is NaCN

Therfore 0.102 moles of NaCN react with 0.102/8 = 0.0127 moles of O2 and excess of Au and water to produce 0.102/2 = 0.051 moles of NaAu(CN)2

=> 0.051 moles of NaAu(CN)2 implies 0.051*272 = 13.877 gms

Therfore, theoretical yield of NaAu(CN)2 = 13.877 gms

b) %yield = (1.2/13.877) *100 = 8.647

c) moles of O2 left = 0.156 - 0.0127 = 0.1433

Amount of O2 left = 0.1433*32 = 4.5856 gms

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