A hole is formed on 0.1 mm thick aluminium sheet on pouring of 1 mL 12 (M) HCl. If the HCl is completely used, then find the area `(cm^(2))` of the hole. (density of Al is `2.7g//cm^(3)`).

Explanation

HCl in 1 mL 12(M) HCl solution
=12 millimol=`12xx10^(-3)`mol
`2Al(2mol=54g)+6HCl(6mol)to2AlCl_(3)+3H_(2)`
Mass of Al dissolved by `12xx10^(-3)` mol of HCl
`=(54xx12xx10^(-3))/(6)g`
Let, the area of the hole on the Al-sheet be `xcm^(2)`.
`therefore`Volume of al dissolved by HCl`=(0.01xx x)cm^(3)` and mass of this amount of `Al=(0.1xx x xx 2.7)g`
`therefore 0.01xx x xx 2.7=(54xx12xx10^(-3))/(6)`
or, `x=(54xx12xx10^(-3))/(6xx0.01xx2.7)=4cm^(2)`.