Question

One process used in gold mines to recover gold from mined rock is as shown below:

___Au + ___NaCN + ___O₂ + ___H₂O → ___NaAu(CN)₂ + ___NaOH

a) Balance the reaction

b) A small test batch is run in 10 mL water. (10 mL of water is 10 g of water) the sample in the reaction contains .0178 g of gold. It is treated with .0075 g of NaCN. O₂ is present in excess (it being in the air). What is the limiting reagent in test reaction? (show calculations)

c) What mass of NaAu(CN)₂ would be produced by the reaction described in question b?

Answer 1

a) 4Au + 8NaCN + 1O2 + 2H2O ----> 4NaAu(CN)2 + 4NaOH

b) H2O moles = mass / molar mass of H2O = 10 /18 = 0.556

gold moles = 0.0178/196.966 = 0.00009

NaCN moles = 0.0075/49 = 0.000153

as per reaction for 0.00009 moles Au the NaCN moles required = ( 8/4) Au moles = ( 8/4) x 0.00009 = 0.00018

but we had only 0.000153 moles NaCN , hence NaCN is limiting reagnet

c) NaAu(CN)2 moles = ( 4/8) NaCN moles = ( 4/8) x 0.000153 = 0.0000765

mass of NaAu(CN)2 = moles x molar mass = 0.0000765 x 272 = 0.02 g