The ore grade is important in determining the metallurgical route for mineral recovery. Aqua regia is one of the classical analytical methods used to determine ore grade. You are tasked to apply the technique to the copper-bearing ore to determine the head grade. The optimized method is to digest a sample corresponding to 3% wt/v of pulp (solid + aqua regia solution) aqua regia solution at 60℃ for 1 hour. a) Define the following: (a) aqua regia, (b) head grade. b) Show the preparation of the aqua regia solution step by step, knowing that the digestion should be done in 150 mL of the aqua regia solution and the concentration of the reagents must be 3.0 M. The available chemicals in the storeroom with their characteristics are listed here below: • HCl; Molecular mass: 36.46 g/mol; Density: 1.28 Kg/L; Purity: 35% • HNO3; Molecular mass: 63.01 g/mol; Density: 1.54 Kg/L; Purity: 60% • H2SO4; Molecular mass: 98.08 g/mol; Density: 1.84 Kg/L; Purity 98% • NaOH (Atomic mass of Na= 22.9 and O= 15.9, and H=1 [g/mol] and Density of NaOH: 2,13 g/cm³). • CuSO4⋅5H2O (Atomic mass of Cu= 63.5, S=32, O= 15.9, and H=1 [g/mol]) c) After digestion, the solution was analyzed for Cu. Results revealed that the solution contains 1500 ppm of the analyte. What is the value of the head grade in %?

Respuesta :

a)
(a) Aqua regia is a highly corrosive mixture of concentrated nitric acid (HNO3) and hydrochloric acid (HCl), typically in a molar ratio of 1:3. It is used in chemical analysis and in the extraction of gold and platinum group metals.

(b) Head grade refers to the average grade of ore fed into a metallurgical process. It is usually expressed as a percentage of the metal content in the ore.

b) Preparation of 150 mL of 3.0 M aqua regia solution:

Step 1: Calculate the volume of each acid needed.
- Molarity (M) = moles of solute / liters of solution
- Given: Volume of aqua regia solution = 150 mL = 0.15 L
- Moles of HCl needed:
Moles = Molarity × Volume (L)
Moles = 3.0 M × 0.15 L = 0.45 moles
- Moles of HNO3 needed:
Moles = Molarity × Volume (L)
Moles = 3.0 M × 0.15 L = 0.45 moles

Step 2: Calculate the mass of each acid needed.
- Mass = Moles × Molar mass
- Mass of HCl needed:
Mass = 0.45 moles × 36.46 g/mol = 16.41 g
- Mass of HNO3 needed:
Mass = 0.45 moles × 63.01 g/mol = 28.35 g

Step 3: Prepare the aqua regia solution.
- Add 16.41 g of 35% HCl to a beaker.
- Add 28.35 g of 60% HNO3 to the same beaker.
- Dilute the solution to 150 mL with water.

c) Calculation of head grade:
- Given: 1500 ppm of Cu in the solution
- ppm (parts per million) = (mass of solute / mass of solution) × 10^6
- Mass of Cu = 1500 ppm × 150 g = 0.15 g
- Head grade = (Mass of Cu / Mass of sample) × 100
= (0.15 g / 5 g) × 100
= 3.0%

Therefore, the head grade of the copper-bearing ore is 3.0%.