Question

4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) → 4 NaAu(CN)2(aq) + 4 NaOH(aq) (2) 2 NaAu(CN)2(aq) + Zn(s) → 2 Au(s) +Na2[Zn(CN)4](aq) If 23 kilograms of ore is 0.19% gold by mass, how much Zn is needed to react with the gold in the ore? Assume that both reactions are 100% efficient.

Answer 1

72.57 grams

Step-by-step explanation:

The mass percentage of gold in the ore is 0.19 %

The mass of ore used = 23 kg

The mass of gold in the ore =
`(0.19X23)/(100)=0.0437kg`

moles of gold in the given mass of ore =
`(mass)/(Atomic mass)=(0.437X1000)/(197)=2.22mol`

As per given equation four moles of Au is giving four moles of complex compound on reacting with NaCN

Then in second reaction two moles of the complex is reacting with one mole of Zinc

Thus two moles of gold are reacting with one mole of Zinc

The moles of Zinc needed = 0.5 X 2.22 mol = 1.11 moles

The mass of Zinc needed = moles X atomic mass =1.11 X 65.38 = 72.57 grams