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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