Problem 58 - Chemistry The Central Science

Question

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 \(31,150\) L of \(\mathrm{H}_{2}\) . He generated the \(\mathrm{H}_{2}\) using the reaction between iron and hy- drochloric 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 \(22^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

Approximately \(71.77\) kilograms of iron were needed to produce the given volume of hydrogen gas at a temperature of \(22^\circ C\).

Step 1: Calculate the moles of hydrogen gas needed

Using the Ideal Gas Law, we can calculate the amount of hydrogen gas in moles that is needed to fill the volume of \(31,150\) L at \(22^\circ C\). The Ideal Gas Law is given by: \[PV=nRT\] where: \(P\) = pressure (assumed to be standard conditions, so \(1 \;atm\)) \(V\) = volume of hydrogen gas ($31,150 \; L\) \(n\) = amount of gas in moles \(R\) = gas constant (\(0.0821 \frac{L \cdot atm}{mol \cdot K}\)) \(T\) = temperature in Kelvin (22°C = \((22+273.15) \; K = 295.15 \; K\)) Rearranging to solve for n, we get: \[n = \frac{PV}{RT}\]

Step 2: Calculate the moles of hydrogen gas

Now, we can plug in the given values and calculate the moles of hydrogen gas needed: \[n = \frac{(1 \; atm)(31,150 \; L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(295.15 \; K)}\] \[n = 1284.29 \; mol\, \ H_{2}\]

Step 3: Use stoichiometry to find the moles of iron needed

From the balanced chemical equation, we have: \[\mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g)\] Since one mole of hydrogen gas is produced per mole of iron reacted, the moles of iron needed is equal to the moles of hydrogen gas. Therefore, we need \(1284.29\) moles of iron.

Step 4: Calculate the mass of iron needed

We can now use the molar mass of iron to calculate the mass of iron needed. The molar mass of iron is \(55.845 \frac{g}{mol}\). Mass of iron = moles of iron × molar mass of iron Mass of iron = \(1284.29 \; mol \times 55.845 \frac{g}{mol}\) Mass of iron = 71768.08 g Converting grams to kilograms: Mass of iron = 71.77 kg So, approximately \(71.77\) kilograms of iron were needed to produce the given volume of hydrogen gas at a temperature of \(22^\circ C\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law, expressed as \( PV = nRT \), is a crucial piece of the puzzle when trying to understand the behavior of gases under various conditions of pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount in moles (\(n\)). It's an equation that connects all these variables, providing a way to solve for one when the others are known.

Let's explore this concept with an example: Balloonist Jacques Charles needed to fill his balloon with hydrogen gas (\(H_2\)), and he had a specific volume in mind. To find out how much gas is needed in moles, the Ideal Gas Law comes into play. By inputting the known values—volume of gas, atmospheric pressure, temperature converted to Kelvin, and the ideal gas constant (\(R\))—we can rearrange the equation to solve for the number of moles of hydrogen gas required.

An often overlooked aspect of the Ideal Gas Law is that the conditions must be

Chemical Reaction Stoichiometry

As students delve deeper into chemistry, they learn about chemical reaction stoichiometry, which allows them to predict the quantities of reactants and products in a chemical reaction. It's based on the law of conservation of mass and the concept that atoms are neither created nor destroyed in chemical reactions.

In the context of our exercise, after using the Ideal Gas Law to determine the moles of hydrogen gas, we utilize stoichiometry to find the equivalent moles of iron consumed in the reaction. The balanced chemical equation provides the ratio of reactants to products, indicating how many moles of hydrogen gas are produced per mole of iron. By understanding stoichiometry, we make sense of the fact that it is a one-to-one ratio for iron and hydrogen gas in this specific reaction.

This is crucial information when calculating the required mass of iron, as it enables us to directly convert the moles of hydrogen gas to moles of iron. This demonstrates stoichiometry's role in bridging the gap between chemical equations and practical applications like filling a balloon with gas.

Molar Mass

Molar mass, the mass of one mole of a substance, is a fundamental concept in chemistry that connects the microscopic world of atoms to the macroscopic world we can measure. It is expressed in grams per mole (g/mol) and can be found on the periodic table for each element.

The exercise presented here is an excellent instance of how molar mass allows chemists to translate moles, an abstract concept, into something tangible like mass. Once the stoichiometry step has been completed, we know how many moles of iron are needed and we simply multiply this number by the molar mass of iron to determine the weight in grams.

This calculation is essential for lab work, where precise amounts of chemicals are necessary. By using the molar mass of iron (55.845 g/mol), we can easily convert our theoretical moles into a weight, allowing us to measure out the exact amount of iron needed in the lab, or in this case, for filling up a balloon with hydrogen gas. Remember, accurate conversions between mass and moles via molar mass are at the heart of successfully carrying out chemical reactions and experiments.