A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by same current in 2 hours if current efficiency is 50%?

Options

A. 22 kg
B. 18 kg
C. 9 kg
D. 27 kg

Explanation

To solve the problem step by step, we can use Faraday's laws of electrolysis, which state that the weight of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step 1: Establish the relationship
According to Faraday's law, we can write the relationship between the weights of the metals produced:
\[ \frac{\text{Weight of Calcium}}{\text{Equivalent Weight of Calcium}} = \frac{\text{Weight of Aluminium}}{\text{Equivalent Weight of Aluminium}} \]

### Step 2: Identify the weights and equivalent weights
- The weight of calcium produced is given as 40 kg.
- The equivalent weight of calcium can be calculated as:
\[ \text{Equivalent Weight of Calcium} = \frac{\text{Molecular Weight}}{\text{Valence}} = \frac{40}{2} = 20 \text{ kg/equiv} \]

- For aluminium, the equivalent weight is:
\[ \text{Equivalent Weight of Aluminium} = \frac{27}{3} = 9 \text{ kg/equiv} \]

### Step 3: Substitute the values into the equation
Now substituting the known values into the relationship:
\[ \frac{40 \text{ kg}}{20 \text{ kg/equiv}} = \frac{\text{Weight of Aluminium}}{9 \text{ kg/equiv}} \]

### Step 4: Solve for the weight of aluminium
Cross-multiplying gives:
\[ 40 \times 9 = 20 \times \text{Weight of Aluminium} \]
\[ 360 = 20 \times \text{Weight of Aluminium} \]
\[ \text{Weight of Aluminium} = \frac{360}{20} = 18 \text{ kg} \]

### Step 5: Adjust for current efficiency
Since the current efficiency is given as 50%, we need to adjust the weight of aluminium produced:
\[ \text{Actual Weight of Aluminium} = \frac{50}{100} \times 18 \text{ kg} = 9 \text{ kg} \]

### Final Answer
Thus, the amount of aluminium that can be produced is **9 kg**.

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