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Chapter 4: Problem 105

A drop \((0.05 \mathrm{mL})\) of \(12.0 \mathrm{M} \mathrm{HCl}\) is spread over a sheet of thin aluminum foil. Assume that all the acid reacts with, and thus dissolves through, the foil. What will be the area, in \(\mathrm{cm}^{2}\), of the cylindrical hole produced? (Density of \(\mathrm{Al}=2.70 \mathrm{g} / \mathrm{cm}^{3} ;\) foil thickness \(=0.10 \mathrm{mm} .)\) \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

Short Answer

Expert verified
The area of the cylindrical hole produced in the aluminum foil is 0.2 cm².

Step by step solution

01

Calculate the number of moles of HCl

Molarity (M) is given by the number of moles of solute per litre of solution. Given that, 12.0 M HCl means there are 12.0 moles of HCl in 1 liter or 1000 mL of solution.So, the number of moles of HCl in 0.05 mL can be calculated using the equation: \[\text{{number of moles of HCl}} = \text{{Molarity}} \times \text{{Volume in litres}}\]Or,\[\text{{number of moles of HCl}} = 12.0M \times 0.05 mL \times \frac{1L}{1000 mL}= 0.0006 \text{{moles}}\]
02

Use stoichiometry to determine the volume of Al

Looking at the balanced chemical equation: \[2Al(s) + 6HCl(aq) \rightarrow 2AlCl3(aq) + 3H2(g)\]we can see that 6 moles of HCl react with 2 moles of Al. Hence 0.0006 moles of HCl will react with:\[\text{{moles of Al}} = \frac{2}{6} \times 0.0006 \text{{ moles of HCl}} = 0.0002 \text{{ moles of Al}}\]Now, using density formula, \(density = \frac{mass}{volume}\), the volume of Al can be calculated as:\[\text{{volume of Al}} = \frac{\text{{moles of Al}} \times \text{{molar mass of Al}}}{\text{{density of Al}}} \]Substituting the values, \[\text{{volume of Al}} = \frac{0.0002 \text{{moles of Al}} \times 26.98 \text{{g/mol}}}{2.70 \text{{g/cm}}^{3}} = 0.002 \text{{cm}}^{3}\]
03

Calculate the area of the hole

The aluminum foil is shown as a cylinder so the area of the hole (base of the cylinder) can be found using volume formula \(V = πr^{2}h\), where r is the radius and h is the height (thickness of the foil). Here, volume \(V = 0.002 \text{{cm}}^{3}\) and thickness \(h = 0.10 \text{{mm}} = 0.01 \text{{cm}}\).So, the area of the hole can be calculated as: \[Area = \frac{V}{h} = \frac{0.002 \text{{cm}}^{3}}{0.01 \text{{cm}}} = 0.2 \text{{cm}}^{2}\]

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Most popular questions from this chapter

In many communities, water is fluoridated to prevent tooth decay. In the United States, for example, more than half of the population served by public water systems has access to water that is fluoridated at approximately \(1 \mathrm{mg} \mathrm{F}^{-}\) per liter. (a) What is the molarity of \(\mathrm{F}^{-}\) in water if it contains \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter? (b) How many grams of solid KF should be added to a \(1.6 \times 10^{8}\) L water reservoir to give a fluoride concentration of \(1.2 \mathrm{mg} \mathrm{F}^{-}\) per liter?

A sulfide of iron, containing \(36.5 \%\) S by mass, is heated in \(\mathrm{O}_{2}(\mathrm{g}),\) and the products are sulfur dioxide and an oxide of iron containing \(27.6 \%\) O, by mass. Write a balanced chemical equation for this reaction.

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

How ma475.15 grams \(\mathrm{HCl}\)ny grams of HCl are consumed in the reaction of \(425 \mathrm{g}\) of a mixture containing \(35.2 \% \mathrm{MgCO}_{3}\) and \(64.8 \% \mathrm{Mg}(\mathrm{OH})_{2},\) by mass? $$\begin{array}{c} \mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{MgCO}_{3}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}(\mathrm{g}) \end{array}$$
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