Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 58

Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783 , Jacques Charles used a balloon that contained approximately \(31,150 \mathrm{~L}\) of \(\mathrm{H}_{2}\). He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hydrochloric acid: $$\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ How many kilograms of iron were needed to produce this volume of \(\mathrm{H}_{2}\) if the temperature was \(22{ }^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The mass of iron needed to produce the given volume of \(H_2\) at the given temperature can be determined through the following steps: 1. Calculate the moles of \(H_2\) using the ideal gas law equation considering atmospheric pressure. 2. Determine the moles of iron required using stoichiometry, which is equal to the moles of \(H_2\). 3. Convert moles of iron into mass (in kg) by multiplying by the molar mass of iron (55.845 g/mol). After following these steps and calculations, the mass of iron required to generate \(31,150 L\) of \(H_2\) at \(22{ }^{\circ} \mathrm{C}\) is approximately \(17.23 kg\).

Step by step solution

01

Determine moles of \(H_2\) using the ideal gas law equation

First, we will need to use the ideal gas law equation, which is: \[PV = nRT\] Where: P = pressure (in atm) V = volume (in L) n = moles of the gas R = ideal gas constant (\(0.0821 \frac{\text{L atm}}{\text{mol K}}\)) T = temperature (in Kelvin) In this problem, the volume \(V = 31,150 L\), the temperature \(T = 22°C = 295 K\), and we assume the pressure to be atmospheric, \(P = 1 atm\). We need to find n, the moles of \(H_2\). Rearranging the equation, we get: \[n = \frac{PV}{RT}\]
02

Calculate the moles of \(H_2\)

Substitute the given values and the assumption of atmospheric pressure into the equation from step 1: \[n = \frac{(1\,\text{atm})(31,150\, \text{L})}{(0.0821\, \frac{\text{L atm}}{\text{mol K}})(295\, \text{K})}\] Now, solve the equation to get the moles of \(H_2\).
03

Use stoichiometry to determine moles of iron required

The balanced equation for the given reaction is: \[\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)\] We can see that \(1\) mole of iron reacts to produce \(1\) mole of \(H_2\). So, the moles of iron required are equal to the moles of \(H_2\) calculated in step 2.
04

Convert moles of iron into mass

To convert moles of iron into mass (in kg), we multiply moles by the molar mass of iron: \[mass\: of\: iron = (moles\: of\: iron)(molar\: mass\: of\: iron)\] where the molar mass of iron = \(55.845 g/mol\). Now multiply the moles of iron we found in step 2 by the molar mass of iron and divide by 1000 to convert the grams into kilograms.
05

Calculate the mass of iron in kilograms

Using the moles of iron from step 2 and the molar mass of iron, calculate the mass of iron needed in kilograms. This will be the final answer.

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.

Start your free trial

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

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

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for $1.0 \mathrm{~L}$ of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of $1.0 \mathrm{~L} ;$ in other words, rate is the amount that diffuses over the time it takes to diffuse.)

A fixed quantity of gas at \(25^{\circ} \mathrm{C}\) exhibits a pressure of $99 \mathrm{kPa}\( and occupies a volume of \)4.00 \mathrm{~L}$ (a) Calculate the volume the gas will occupy if the pressure is increased to \(202.6 \mathrm{kPa}\) while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(100^{\circ} \mathrm{C}\) while the pressure is held constant.

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\mathrm{mfp}}\), like the ideal-gas constant) and define units for \(R_{\mathrm{mfp}^{-}}\)

Consider a mixture of two gases, \(\mathrm{A}\) and \(\mathrm{B}\), confined in a closed vessel. A quantity of a third gas, \(\mathrm{C}\), is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,(\mathbf{b})\) the total pressure in the vessel, (c) the mole fraction of gas B?

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4},\) is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is \(1 \mathrm{ppb}\) (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and 101.3 kPa pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation
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