Chemisty

Relative atomic masses:Au= 197, Na = 23, C = 12, N = 14, Zn = 65.4

a.
4Au(s)+ 8NaCN(aq) + O2(g) + 2H2O(l) → 4NaAu(CN)2(aq) + 4NaOH (aq)
Massof Au = 1 x 1000 x 1000 x 0.009% = 90 g
No. of moles of Au = 90/197 mol
No. of moles of NaCN needed = (90/197) x (8/4) = 180/197 mol
Mass of NaCN needed = (180/197) x (23 + 12 + 14) = 44.8 g


b.
4Au(s)+ 8NaCN(aq) + O2(g) + 2H2O(l) → 4NaAu(CN)2(aq) + 4NaOH (aq)
No.of moles of Au = 90/197 mol
No. of moles of NaAu(CN)2 = 90/197 mol

2NaAu(CN)2(aq)+ Zn(s) → 2Au(s) + Na2[Zn(CN)4](aq)
No.of moles of NaAu(CN)2 = 90/197 mol
No. of moles of Zn needed = (90/197) x (1/2) = 45/197 mol
Mass of Zn needed = (45/197) x 65.4 = 14.9g


c.
Mass of Au recovered = 90 g
Volume of Au recovered = 90/19.3 cm³
The length of one side of the cube = ³√(90/19.3) = 1.67cm

2011-07-23 17:48:04 補充：
a.
Mass of Au = Total mass x percentage

No.of moles of Au = mass / molar mass = .....

Mole ratio Au : NaCN = 4 : 8
No. of moles of NaCN = .....

Mass of NaCN = No. of moles x molar mass = .....

2011-07-23 17:48:37 補充：
b.
1 mol of Au yield 1 mol of NaAu(CN)2,
No. of moles of NaAu(CN)2 = No. of moles of Au = .....

From the 2nd equation, mole ratio NaAu(CN)2: Zn = 2 : 1
No. of moles of Zn needed = .....

Mass of Zn needed = No. of moles x molar mass = .....

2011-07-23 17:48:54 補充：
c.
Mass of Au recovered = 90 g

Volume of Au recovered = mass/density = .....

The length of one side of the cube = ³√density = ......