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A 1.65 g sample of \(\mathrm{Al}\) reacts with excess \(\mathrm{HCl}\), and the liberated \(\mathrm{H}_{2}\) is collected over water at \(25^{\circ} \mathrm{C}\) at a barometric pressure of \(744 \mathrm{mmHg} .\) What volume of gaseous mixture, in liters, is collected? $$2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})$$

Short Answer

Expert verified
The corrected volume of the gaseous mixture collected is 2.42 L.

Step by step solution

01

Calculate moles of Al

First, compute for the moles of Al using the given mass and the molar mass of Al, which is \( 26.982 \, g/mol\). Moles of Al \( = \frac{1.65 \, g}{26.982 \, g/mol} = 0.0611 \, mol\)
02

Calculate moles of H2

Determine the moles of \( H_{2} \) using the stoichiometric ratio in the balanced chemical reaction, \( 2Al:3H_{2} = 0.0611 \, mol \, Al:x \, mol \, H_{2}\). Solve for \( x \), getting \( x = \frac{3}{2} \times 0.0611 \, mol \, Al = 0.0916 \, mol \, H_{2}\).
03

Determine the gas volume using the Ideal Gas Law

Use the Ideal Gas Law \(PV = nRT\) where \(P\) is the pressure in atm \(= 744 \, mmHg \times \frac{1 \, atm}{760 \, mmHg} = 0.9789 \, atm\), \(n\) is the number of moles \(= 0.0916 \, mol\), \(R\) is the gas constant \(= 0.0821 \, L.atm/(mol.K)\), and \(T\) is the temperature in Kelvin \(= 25^{\circ} C + 273.15 = 298.15 \, K\). Solving for \(V\), \(V = \frac{nRT}{P}\) gets \(V = \frac{0.0916 \, mol \times 0.0821 \, L.atm/(mol.K) \times 298.15 \, K }{0.9789 \, atm} = 2.30 \, L\).
04

Correct the collected volume for vapor pressure

Lastly, correct for the presence of water vapor in the collected gas. The partial pressure of hydrogen is given by subtracting the water vapor pressure from the total pressure. The vapor pressure of water at \(25^{\circ}C\) is 23.76 mmHg or 0.0313 atm. Hence, the corrected volume of \(H_{2}\) gas is \(V' = nRT/P' = \frac{0.0916 \, mol \times 0.0821 \, L.atm/(mol.K) \times 298.15 \, K }{0.9789 \, atm - 0.0313 \, atm} = 2.42 \, L\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the study of quantitative relationships in chemical reactions. It involves using the balanced chemical equation to determine the proportions of reactants and products. In this exercise, stoichiometry helps us convert the amount of aluminum (Al) into hydrogen gas (\(H_{2}\)) produced. The balanced equation tells us that 2 moles of Al produce 3 moles of \(H_{2}\):
  • Start with the given mass of Al.
  • Convert mass to moles using Al's molar mass.
  • Use the mole ratio from the balanced equation to find moles of \(H_{2}\).
This process bridges the gap between the amount of reactants and products, allowing precise predictions in chemical reactions.
Ideal Gas Law
The Ideal Gas Law links the physical properties of gases using the formula \(PV = nRT\). It helps determine the behavior of gases under different conditions of pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and moles (\(n\)). In this problem, this law helps calculate the volume of hydrogen gas collected.
  • First, convert all units to match: pressure to atm, and temperature to Kelvin.
  • Insert the number of moles of hydrogen gas calculated from stoichiometry.
  • Use the gas constant \(R = 0.0821\, \text{L.atm/(mol.K)}\).
Solving for the unknown volume \(V\), we find the initial gas volume before adjusting for water vapor.This calculation gives us insight into how gasses behave and how the physical conditions affect this behavior.
Chemical Reactions
Chemical reactions describe how atoms are reorganized, resulting in products with new properties. They are represented by balanced equations that maintain the conservation of mass and atoms. In this exercise, we examine the reaction between aluminum (\(2Al\)) and hydrochloric acid (\(6HCl\)), producing aluminum chloride (\(2AlCl_{3}\)) and hydrogen gas (\(3H_{2}\)).
  • Balanced equations ensure the same number of each type of atom on both sides.
  • Determine which compounds are reactants and products.
  • Predict the reaction outcomes using the balanced equation.
This reaction particularly involves single replacement, where hydrogen in hydrochloric acid is displaced by aluminum, showcasing the rearrangement of atoms into different substances. Understanding reactions further helps us control and predict outcomes in various chemical processes.

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Most popular questions from this chapter

A 1.072 g sample of \(\mathrm{He}(\mathrm{g})\) is found to occupy a volume of 8.446 L when collected over hexane at \(25.0^{\circ} \mathrm{C}\) and \(738.6 \mathrm{mmHg}\) barometric pressure. Use these data to determine the vapor pressure of hexane at \(25^{\circ} \mathrm{C}\).

What is the pressure, in pascals, exerted by \(1242 \mathrm{g}\) CO(g) when confined at \(-25^{\circ} \mathrm{C}\) to a cylindrical tank \(25.0 \mathrm{cm}\) in diameter and \(1.75 \mathrm{m}\) high?

An 886 mL sample of \(\mathrm{Ne}(\mathrm{g})\) is at \(752 \mathrm{mmHg}\) and \(26^{\circ} \mathrm{C}\). What will be the new volume if, with the pressure and amount of gas held constant, the temperature is (a) increased to \(98^{\circ} \mathrm{C} ;\) (b) lowered to \(-20^{\circ} \mathrm{C} ?\)

The amount of ozone, \(\mathrm{O}_{3}\), in a mixture of gases can be determined by passing the mixture through a solution of excess potassium iodide, KI. Ozone reacts with the iodide ion as follows: $$\begin{aligned} \mathrm{O}_{3}(\mathrm{g})+3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) & \longrightarrow \\ \mathrm{O}_{2}(\mathrm{g})+\mathrm{I}_{3}^{-}(\mathrm{aq}) &+2 \mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$ The amount of \(I_{3}^{-}\) produced is determined by titrating with thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}:\) $$\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \longrightarrow 3 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})$$ A mixture of gases occupies a volume of \(53.2 \mathrm{L}\) at \(18^{\circ} \mathrm{C}\) and \(0.993 \mathrm{atm} .\) The mixture is passed slowly through a solution containing an excess of KI to ensure that all the ozone reacts. The resulting solution requires \(26.2 \mathrm{mL}\) of \(0.1359 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) to titrate to the end point. Calculate the mole fraction of ozone in the original mixture.

If the Kelvin temperature of a sample of ideal gas doubles (e.g., from 200 K to 400 K), what happens to the root-mean-square speed, \(u_{\mathrm{rms}}\) ? (a) \(u_{\mathrm{rms}}\) increases by a factor of \(\sqrt{2} ;\) (b) \(u_{\mathrm{rms}}\) increases by a factor of \(2 ;(\mathrm{c}) u_{\mathrm{rms}}\) decreases by a factor of 2 (d) \(u_{\mathrm{rms}}\) increases by a factor of \(4 ;\) (e) \(u_{\mathrm{rms}}\) decreases by a factor of 4.

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