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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.

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Key Concepts

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

Stoichiometry
Stoichiometry helps us understand the quantitative relationships in chemical reactions. It's like following a recipe, ensuring you have the right amount of ingredients for your dish. In chemical equations, stoichiometry tells us how much of each reactant is needed to form a desired amount of product.
In the exercise's reaction:
  • One mole of iron (\(\mathrm{Fe}\)) reacts with two moles of hydrochloric acid (\(\mathrm{HCl}\)).
  • This produces one mole of \(\mathrm{H_2}\) gas.
It’s crucial to balance the chemical equation first, as this shows the exact proportions needed.
For example, producing 31,150 liters of \(\mathrm{H_2}\) requires using the Ideal Gas Law first to find moles of \(\mathrm{H_2}\) and then using stoichiometry to connect it back to moles of iron. This ensures each component is measured correctly, much like making sure you don’t have too much or too little of any ingredient in a recipe.
Mole Concept
The mole concept is a way to count particles like atoms and molecules in chemistry. It gives us a bridge between the atomic scale and the real-world scale by using Avogadro’s number, which is \(6.022 \times 10^{23}\) particles per mole.
In the context of the exercise, you start by calculating the moles of \(\mathrm{H_2}\) gas using the ideal gas law: \(PV = nRT\).
Here:
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant (\(0.0821 \text{ L atm/mol K}\)).
By inputting the volume, temperature, and pressure, you can solve for \(n\), giving you the moles of \(\mathrm{H_2}\).
This understanding enables you to convert between mass and moles, moving from the abstract atomic world to measurable quantities. Knowing how to manipulate these calculations is key in any chemistry task involving gases.
Chemical Reactions
Chemical reactions are transformations where substances change into new substances. This is shown using a balanced chemical equation. For the exercise, when iron reacts with hydrochloric acid:
  • Iron (\(\mathrm{Fe}\)) and hydrochloric acid (\(\mathrm{HCl}\)) are the reactants.
  • Iron chloride (\(\mathrm{FeCl_2}\)) and hydrogen gas (\(\mathrm{H_2}\)) are the products.
Every reaction involves the breaking of bonds in reactants and forming new bonds in products.
The balanced equation \(\mathrm{Fe}(s) + 2\,\mathrm{HCl}(aq) \rightarrow \mathrm{FeCl}_2(aq) + \mathrm{H}_2(g)\) shows the proportionate amounts of each substance involved.
This understanding is crucial because it reveals not just what substances will be formed but in what amounts, guiding you to connect observable changes back to molecular interactions.

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Most popular questions from this chapter

A set of bookshelves rests on a hard floor surface on four legs, each having a cross-sectional dimension of \(4.0 \times 5.0 \mathrm{~cm}\) in contact with the floor. The total mass of the shelves plus the books stacked on them is \(200 \mathrm{~kg}\). Calculate the pressure in atmospheres exerted by the shelf footings on the surface.

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), can be prepared by the reaction of calcium carbide with water: $$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$$ Calculate the volume of \(\mathrm{C}_{2} \mathrm{H}_{2}\) that is collected over water at \(23^{\circ} \mathrm{C}\) by reaction of \(1.524 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) if the total pressure of the gas is \(100.4 \mathrm{kPa}\). (The vapor pressure of water is tabulated in Appendix B.)

If \(5.15 \mathrm{~g}\) of \(\mathrm{Ag}_{2} \mathrm{O}\) is sealed in a \(75.0-\mathrm{mL}\) tube filled with 101.3 \(\mathrm{kPa}\) of \(\mathrm{N}_{2}\) gas at \(32{ }^{\circ} \mathrm{C},\) and the tube is heated to \(320^{\circ} \mathrm{C}\), the \(\mathrm{Ag}_{2} \mathrm{O}\) decomposes to form oxygen and silver. What is the total pressure inside the tube assuming the volume of the tube remains constant?

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and \(99.3 \mathrm{kPa}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

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