A drop `(0.05 mL)` of `12 M HCl` is spread over a thin sheet of aluminium foil ( thickness `0.10mm` and density of `Al=2.70g//mL`). Assuming whole of `HCl` is used to dissolve `Al`, what will be maximum area of hole produced in foil ?

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the amount of HCl in moles
Given:
- Volume of HCl = 0.05 mL
- Molarity of HCl = 12 M

To find the number of moles of HCl, we use the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume (in L)} \]
Convert volume from mL to L:
\[ 0.05 \, \text{mL} = 0.05 \times 10^{-3} \, \text{L} = 5 \times 10^{-5} \, \text{L} \]
Now calculate the moles:
\[ \text{Moles of HCl} = 12 \, \text{mol/L} \times 5 \times 10^{-5} \, \text{L} = 6 \times 10^{-4} \, \text{mol} \]

### Step 2: Determine the equivalent mass of Al that can be dissolved
The reaction between HCl and Al can be represented as:
\[ 2 \, \text{HCl} + \text{Al} \rightarrow \text{AlCl}_3 + \text{H}_2 \]
From the reaction, 2 moles of HCl react with 1 mole of Al. Therefore, the moles of Al that can be dissolved by the given moles of HCl is:
\[ \text{Moles of Al} = \frac{\text{Moles of HCl}}{2} = \frac{6 \times 10^{-4}}{2} = 3 \times 10^{-4} \, \text{mol} \]

### Step 3: Calculate the mass of Al that can be dissolved
The molar mass of Al is approximately 27 g/mol. Therefore, the mass of Al that can be dissolved is:
\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar mass of Al} = 3 \times 10^{-4} \, \text{mol} \times 27 \, \text{g/mol} = 0.0081 \, \text{g} \]

### Step 4: Calculate the volume of Al that corresponds to this mass
Using the density of Al (2.70 g/mL), we can find the volume:
\[ \text{Volume of Al} = \frac{\text{Mass of Al}}{\text{Density of Al}} = \frac{0.0081 \, \text{g}}{2.70 \, \text{g/mL}} \approx 0.0030 \, \text{mL} \]

### Step 5: Calculate the area of the hole produced
Given the thickness of the aluminum foil is 0.10 mm (which is 0.01 cm), we can use the formula for volume:
\[ \text{Volume} = \text{Area} \times \text{Thickness} \]
Rearranging for Area:
\[ \text{Area} = \frac{\text{Volume}}{\text{Thickness}} = \frac{0.0030 \, \text{mL}}{0.01 \, \text{cm}} = 0.30 \, \text{cm}^2 \]

### Final Answer
The maximum area of the hole produced in the foil is approximately:
\[ \text{Area} \approx 0.30 \, \text{cm}^2 \]