Final answer:
To extract gold from 40 kg of rocks with 2% gold, you'd need 40.6 liters of a 0.200 M sodium cyanide solution. This calculation is based on the stoichiometry of the reaction between sodium cyanide and gold.
Step-by-step explanation:
To calculate how many liters of a 0.200 M sodium cyanide solution would be needed to react with 40.0 kg of rocks containing 2.00 % by mass of gold, we must first determine the mass of gold in the rocks, then convert that mass to moles, and finally find the volume of the sodium cyanide solution required using its molarity.
First, calculate the mass of gold in the rocks:
40.0 kg of rocks × 0.02 = 0.8 kg of gold (800 g of gold).
The molar mass of gold (Au) is approximately 197 g/mol, so the number of moles of gold is:
800 g / 197 g/mol ≈ 4.06 mol of Au.
Gold reacts with sodium cyanide (NaCN) following the reaction:
4 Au + 8 NaCN + O₂ + 2 H₂O → 4 Na[Au(CN)₂] + 4 NaOH.
From the reaction equation, we see that each mole of gold requires 2 moles of NaCN. Therefore, the moles of NaCN needed are:
4.06 mol Au × 2 mol NaCN/mol Au = 8.12 mol NaCN.
Finally, using the molarity of the NaCN solution (M = moles/volume), we can find the volume V:
V = moles of NaCN / Molarity of NaCN
V = 8.12 mol NaCN / 0.200 mol/L ≈ 40.6 L.
Therefore, you would need approximately 40.6 liters of a 0.200 M sodium cyanide solution to react with the gold in the rocks.