Question

A drop \[{\text{( 0}}{\text{.05 mL)}}\] of \[{\text{12 M HCl}}\] is spread over a thin sheet of aluminium foil (thickness \[{\text{0}}{\text{.10 mm}}\] and density of \[{\text{Al }} = 2.70{\text{ g/mL }}\]). Assuming whole of the \[{\text{HCl}}\] is used to dissolve \[{\text{Al}}\], the maximum area of the hole produced in the foil will be ( write it as \[0.6 = 6\] (only one digit)):

Answer 1

Hint: Number of moles of \[{\text{HCl}}\] can be calculated using the molarity and volume. The number of moles of \[{\text{Al}}\] will be the same as calculated of \[{\text{HCl}}\]. We need to calculate the mass of aluminium and the volume of aluminium using the formula. With the basic relation in volume area and thickness we can calculate the area.
Formula used: \[{\text{Molarity }} = \dfrac{{{\text{no}}{\text{. of moles}}}}{{{\text{volume (in L)}}}}\]
\[{\text{no}}{\text{. of moles }} = \dfrac{{{\text{Mass}}}}{{{\text{molar mass}}}}\]
\[{\text{volume}} = {\text{area}} \times {\text{thickness}}\]

Complete step by step answer:
Using the formula we can calculate the number of mole of \[{\text{HCl}}\] , molarity and volume is given to us:
\[12{\text{ M }} = \dfrac{{{\text{no}}{\text{. of moles}} \times {\text{1000}}}}{{{\text{0}}{\text{.05mL}}}}\]
Rearranging the equation we will get the number of moles:
\[{\text{no}}{\text{. of moles of HCl}} = 0.6 \times {10^{ - 3}}\]
The reaction of aluminium with \[{\text{HCl}}\] yields \[{\text{AlC}}{{\text{l}}_3}\].
Since the entire amount of \[{\text{HCl}}\] is dissolved or reacted with aluminium and 3 moles of chlorine ion joins to 1 mole of aluminium therefore for \[0.6 \times {10^{ - 3}}\] moles \[0.2 \times {10^{ - 3}}\] moles of aluminium will react. This is the simple unitary method we have used.
Using the number of moles we can calculate the mass of aluminum as molar mass of aluminium is \[27{\text{ g mo}}{{\text{l}}^{ - 1}}\]
\[ \Rightarrow 0.2 \times {10^{ - 3}}{\text{ }} = \dfrac{{{\text{Mass}}}}{{27{\text{ g mo}}{{\text{l}}^{ - 1}}}}\]
\[ \Rightarrow {\text{Mass }} = 0.2 \times {10^{ - 3}}{\text{ }} \times 27{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Now we have been given the value of density and mass has been calculated by us. So we will get the mass of aluminium as:
\[{\text{Density}} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}\]
We can rewrite the above equation as:
\[{\text{volume}} = \dfrac{{{\text{mass}}}}{{{\text{Density}}}}\]
Substituting the given and calculated values we will get:
\[ \Rightarrow {\text{volume}} = \dfrac{{0.2 \times {{10}^{ - 3}}{\text{ }} \times 27{\text{ g mo}}{{\text{l}}^{ - 1}}}}{{{\text{2}}{\text{.7g m}}{{\text{l}}^{ - 1}}}} = 2 \times {10^{ - 3}}{\text{mL}}\]
We will use the following relation to find the area:
\[{\text{volume}} = {\text{area}} \times {\text{thickness}}\]
we will rearrange it and write it as:
\[\dfrac{{{\text{volume}}}}{{{\text{thickness}}}} = {\text{area}}\]
\[ \Rightarrow \dfrac{{2 \times {{10}^{ - 3}}{\text{ml}}}}{{0.1{\text{ mm}}}} = {\text{area}}\]
We need to use the following unit conversion:
\[1{\text{ mL }} = 1{\text{c}}{{\text{m}}^3} = {10^3}{\text{m}}{{\text{m}}^3}\]
\[{\text{area}} = \dfrac{{2 \times {{10}^{ - 3}} \times {{10}^3}{\text{m}}{{\text{m}}^2}}}{{0.1{\text{ mm}}}}\]
\[ \Rightarrow {\text{area}} = 20{\text{ m}}{{\text{m}}^2} = 0.2{\text{ c}}{{\text{m}}^2}\]
But we have to report the answer in only 1 digit hence the answer is 2.

Note:
The reaction of aluminium and hydrochloric acid occurs as follows and the products are aluminium chloride and hydrogen liberated as gas.
\[{\text{Al}} + 3{\text{HCl}} \to {\text{AlC}}{{\text{l}}_3} + _2^3{{\text{H}}_2}\]. This equation can be converted into a whole number by multiplying it with 2.