One drop (assume 0.050 mL) of 6.0 M HCl is placed onto the

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