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Chapter 6: Problem 67

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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Most popular questions from this chapter

A 2.00 L container is filled with \(\operatorname{Ar}(g)\) at 752 mm Hg and \(35^{\circ} \mathrm{C} .\) A \(0.728 \mathrm{g}\) sample of \(\mathrm{C}_{6} \mathrm{H}_{6}\) vapor is then added. (a) What is the total pressure in the container? (b) What is the partial pressure of \(\mathrm{Ar}\) and of \(\mathrm{C}_{6} \mathrm{H}_{6} ?\)

Carbon monoxide, \(\mathrm{CO}\), and hydrogen react according to the equation below. $$3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of which reactant gas remains if \(12.0 \mathrm{LCO}(\mathrm{g})\) and \(25.0 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) are allowed to react? Assume that the volumes of both gases are measured at the same temperature and pressure.

A 10.0 g sample of a gas has a volume of \(5.25 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(762 \mathrm{mm} \mathrm{Hg} .\) If \(2.5 \mathrm{g}\) of the same gas is added to this constant 5.25 L volume and the temperature raised to \(62^{\circ} \mathrm{C},\) what is the new gas pressure?

A 0.418 g sample of gas has a volume of \(115 \mathrm{mL}\) at \(66.3^{\circ} \mathrm{C}\) and \(743 \mathrm{mm} \mathrm{Hg} .\) What is the molar mass of this gas?

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?
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