Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 6.11-g sample of a Cu-Zn alloy reacts with \(\mathrm{HCl}\) acid to produce hydrogen gas. If the hydrogen gas has a volume of \(1.26 \mathrm{~L}\) at \(22^{\circ} \mathrm{C}\) and \(728 \mathrm{mmHg},\) what is the percent of Zn in the alloy? (Hint: Cu does not react with HCl.)

Short Answer

Expert verified
The percent of Zn in the alloy is approximately 54.12%.

Step by step solution

01

Identify the Reaction

Understand that Zinc (Zn) reacts with hydrochloric acid (HCl) to produce hydrogen gas (H₂), while Copper (Cu) does not. The reaction is: \[ \mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2 \] This indicates that all produced hydrogen gas is due to the reaction with Zn.
02

Use Ideal Gas Law to Find Moles of Hydrogen

The volume of hydrogen gas produced is given as \(1.26 \text{ L}\) at \(22^{\circ} \text{C}\) and \(728 \text{ mmHg}\). Use the ideal gas law \( PV = nRT \) to find the number of moles of hydrogen, \( n \). Convert the temperature to Kelvin: \( T = 22 + 273.15 = 295.15 \text{ K} \), and pressure to atm: \( P = \frac{728}{760} \text{ atm} = 0.958 \text{ atm} \). \( R \) is the gas constant \(0.0821 \text{ L atm/mol K} \). Substitute values into the equation: \[ n = \frac{PV}{RT} = \frac{0.958 \times 1.26}{0.0821 \times 295.15} \approx 0.0506 \text{ moles of } \mathrm{H}_2 \]
03

Relate Moles of Hydrogen to Moles of Zinc

From Step 1, the equation \( \mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2 \) shows that 1 mole of Zn produces 1 mole of H₂. Therefore, moles of Zn \( = 0.0506 \).
04

Calculate Mass of Zinc

Find the mass of Zn using its molar mass (\(65.38 \text{ g/mol}\)). \[ \text{Mass of Zn} = 0.0506 \times 65.38 = 3.307 \text{ g} \]
05

Determine Percent of Zn in the Alloy

Calculate the percent of Zn in the alloy: \[ \text{Percent of Zn} = \frac{\text{Mass of Zn}}{\text{Total Mass of Alloy}} \times 100 = \frac{3.307}{6.11} \times 100 \approx 54.12\% \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Cu-Zn alloy
A Cu-Zn alloy is a composite material made primarily of copper and zinc. In this particular scenario, the alloy is composed of these two metals, but only the zinc component reacts with hydrochloric acid (HCl).
When mixed, the alloy doesn't behave like a pure metal. Instead, it takes on unique properties from both copper and zinc. This makes it useful in various applications like coins and corrosion-resistant materials.
Notably, in reactions with acids, only the zinc part reacts under these conditions. Copper remains unreactive with HCl. This understanding is crucial to determine the zinc content in the alloy. We measure the reaction involving the alloy to deduce how much zinc it contains.
hydrogen gas
Hydrogen gas (H₂) is a byproduct of the reaction between zinc and hydrochloric acid (HCl). This produces a noticeable release of gas bubbles during the reaction.
In the context of the alloy, when zinc reacts with HCl, hydrogen gas is produced. The volume of hydrogen can be measured under specific conditions of temperature and pressure to understand more about the reaction.
Hydrogen gas is colorless, odorless, and is the simplest element. It's essential to calculate its amount using the volume data from the reaction. This information helps in connecting the dots between the amount of zinc present and the amount of hydrogen gas produced.
ideal gas law
The Ideal Gas Law is a critical equation in chemistry, given by \( PV = nRT \), which relates the pressure, volume, number of moles, and temperature of a gas.
In this exercise, the Ideal Gas Law is applied to determine the number of moles of hydrogen gas produced. Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
By rearranging and substituting appropriate values into the equation, we calculate the moles of hydrogen. This data serves as a bridge to identify the moles, and thus the mass, of zinc that reacted, thanks to its direct stoichiometric relation in this specific reaction.
stoichiometry
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It's all about understanding how the quantities of different substances relate to each other.
In this problem, the stoichiometric equation \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \) plays a vital role. Each mole of zinc reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.
Understanding this relationship allows us to seamlessly connect the moles of hydrogen gas obtained from the Ideal Gas Law to the amount of zinc present. Stoichiometry provides the proportional reasoning necessary for calculating the percent composition of zinc in the alloy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) $$ If \(5.97 \mathrm{~g}\) of glucose reacts and \(1.44 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) gas is collected at \(293 \mathrm{~K}\) and \(0.984 \mathrm{~atm},\) what is the percent yield of the reaction?

Assuming that air contains 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2},\) and 1.0 percent Ar, all by volume, how many molecules of each type of gas are present in \(1.0 \mathrm{~L}\) of air at STP?

A certain anesthetic contains 64.9 percent \(\mathrm{C}, 13.5\) percent \(\mathrm{H}\), and 21.6 percent \(\mathrm{O}\) by mass. At \(120^{\circ} \mathrm{C}\) and \(750 \mathrm{mmHg}, 1.00 \mathrm{~L}\) of the gaseous compound weighs \(2.30 \mathrm{~g}\). What is the molecular formula of the compound?

Calculate the density of helium in a helium balloon at \(25.0^{\circ} \mathrm{C}\). (Assume that the pressure inside the balloon is \(1.10 \mathrm{~atm} .)\)

The apparatus shown here can be used to measure atomic and molecular speeds. Suppose that a beam of metal atoms is directed at a rotating cylinder in a vacuum. A small opening in the cylinder allows the atoms to strike a target area. Because the cylinder is rotating, atoms traveling at different speeds will strike the target at different positions. In time, a layer of the metal will deposit on the target area, and the variation in its thickness is found to correspond to Maxwell's speed distribution. In one experiment it is found that at \(850^{\circ} \mathrm{C}\) some bismuth (Bi) atoms struck the target at a point \(2.80 \mathrm{~cm}\) from the spot directly opposite the slit. The diameter of the cylinder is \(15.0 \mathrm{~cm},\) and it is rotating at 130 revolutions per second. (a) Calculate the speed (in \(\mathrm{m} / \mathrm{s}\) ) at which the target is moving. (Hint: The circumference of a circle is given by \(2 \pi r\), where \(r\) is the radius.) (b) Calculate the time (in seconds) it takes for the target to travel \(2.80 \mathrm{~cm} .\) (c) Determine the speed of the \(\mathrm{Bi}\) atoms. Compare your result in part (c) with the \(u_{\mathrm{rms}}\) of \(\mathrm{Bi}\) at \(850^{\circ} \mathrm{C}\). Comment on the difference.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free