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Question: 15. A drop (0.05 mL) of 12 M HCI is spread over a thin sheet of aluminium foil (thickness 1x$10^{-2}...

  1. A drop (0.05 mL) of 12 M HCI is spread over a thin sheet of aluminium foil (thickness 1x10−210^{-2}cm and density 2.7 gmL−1^{-1}). Assuming whole of the HCI is used to dissolve aluminium, the maximum area of hole produced in the foil is
A

2×10−110^{-1}cm²

B

20x10−110^{-1}cm²

C

200 x 10−110^{-1}cm²

D

2000×10−110^{-1}cm²

Answer

2×10−110^{-1}cm²

Explanation

Solution

Here's a breakdown of the solution:

  1. Calculate moles of HCl: Given volume of HCl = 0.05 mL = 0.05×10−30.05 \times 10^{-3} L Given molarity of HCl = 12 M Moles of HCl = Molarity × Volume (in L) Moles of HCl = 12 mol/L×0.05×10−3 L=0.6×10−3 mol=6×10−4 mol12 \text{ mol/L} \times 0.05 \times 10^{-3} \text{ L} = 0.6 \times 10^{-3} \text{ mol} = 6 \times 10^{-4} \text{ mol}

  2. Write the balanced chemical equation: The reaction between aluminum (Al) and hydrochloric acid (HCl) is: 2Al (s)+6HCl (aq)→2AlCl3 (aq)+3H2 (g)2\text{Al (s)} + 6\text{HCl (aq)} \rightarrow 2\text{AlCl}_3\text{ (aq)} + 3\text{H}_2\text{ (g)} From the stoichiometry, 6 moles of HCl react with 2 moles of Al, which simplifies to 3 moles of HCl reacting with 1 mole of Al.

  3. Determine moles of Al reacted: Moles of Al = (Moles of HCl) / 3 Moles of Al = (6×10−4 mol)/3=2×10−4 mol(6 \times 10^{-4} \text{ mol}) / 3 = 2 \times 10^{-4} \text{ mol}

  4. Calculate mass of Al reacted: Molar mass of Al = 27 g/mol Mass of Al = Moles of Al × Molar mass of Al Mass of Al = 2×10−4 mol×27 g/mol=54×10−4 g=5.4×10−3 g2 \times 10^{-4} \text{ mol} \times 27 \text{ g/mol} = 54 \times 10^{-4} \text{ g} = 5.4 \times 10^{-3} \text{ g}

  5. Calculate volume of Al reacted: Density of Al = 2.7 g/mL Volume of Al = Mass of Al / Density of Al Volume of Al = (5.4×10−3 g)/(2.7 g/mL)=2×10−3 mL(5.4 \times 10^{-3} \text{ g}) / (2.7 \text{ g/mL}) = 2 \times 10^{-3} \text{ mL} Since 1 mL = 1 cm³, Volume of Al = 2×10−3 cm32 \times 10^{-3} \text{ cm}^3

  6. Calculate the area of the hole: The volume of the foil dissolved is equal to the volume of Al reacted. Volume of foil = Area × Thickness Given thickness of foil = 1×10−21 \times 10^{-2} cm Area = Volume of Al / Thickness Area = (2×10−3 cm3)/(1×10−2 cm)(2 \times 10^{-3} \text{ cm}^3) / (1 \times 10^{-2} \text{ cm}) Area = 2×10(−3−(−2)) cm2=2×10−1 cm22 \times 10^{(-3 - (-2))} \text{ cm}^2 = 2 \times 10^{-1} \text{ cm}^2

The maximum area of the hole produced in the foil is 2×10−1 cm22 \times 10^{-1} \text{ cm}^2.