One drop (assume 0.050 mL) of 6.0 M HCl is placed onto the surface of 0.028-mm-thick aluminum foil. What is the maximum diameter of the hole that will result from the HCl dissolving the aluminum? (Density of aluminum = 2.7 g/cm3)

Solution 107P:

Here, we are going to calculate the diameter of the hole that will result from the HCl dissolving the aluminium.

Step1:

Given,

Volume of HCl = 0.050 mL = 0.050 / 1000 L = 0.000050 L [1000 mL = 1 L]

Molarity of HCl = 6.0 M

Therefore, number of moles of HCl = Volume of HCl x Molarity

= 0.000050 L x 6.0 M

= 0.0003 mol

Step2:

The reaction between Al and HCl takes place in the following manner:

2Al + 6HCl → 2AlCl3 + 3H2

From the above equation, it is clear that, 6 mol of HCl dissolves 2 mol of Al.

Therefore, 0.0003 mol of HCl will dissolve (2/6 x 0.0003 = 0.0001) mol of Al

Step3:

1 mol Al = molar mass of Al

Therefore, 0.0001 mol Al = 0.0001 x molar mass of Al

= 0.0001 x 26.98 g/mol

= 0.002698 g

Step4:

Density of Al = 2.7 g/cm3

Therefore, volume of Al reacted = mass of Al / density

= 0.002698 g / 2.7 g cm-3

= 0.001 cm3

Step5:

Since the hole is cylindrical in shape, therefore, volume of the hole is given by = 𝛑r2h, where, r is the radius of the hole and h is the thickness.

Volume of the hole = volume of Al reacted = 0.001 cm3

Thickness of the hole = thickness of the Al foil = 0.028 mm

= 0.028 / 10 cm [10 mm = 1 cm]

= 0.0028 cm

Substituting the values in the formula, we have,

3.14 x r2 x 0.0028 cm = 0.001 cm3

0.008792 cm x r2 = 0.001 cm3

r2 = 0.001 cm3 / 0.008792 cm

r2 = 0.11 cm

r =

r = 0.33 cm (approx.)

Therefore, diameter = 2 x radius

= 2 x 0.33 cm

= 0.66 cm

Thus, the diameter of the hole is 0.66 cm.

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