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%.
(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%.