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Chapter 4: Problem 152

An alloy of iron and carbon was treated with sulfuric acid. in which only iron reacts. \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g)\) If a sample of alloy weighing \(2.358\) g gave \(0.1152 \mathrm{~g}\) of hydro- gen, what is the percentage of iron in the alloy?

Short Answer

Expert verified
The percentage of iron in the alloy is approximately 89.92%.

Step by step solution

01

Determine Moles of Hydrogen Produced

To find the moles of hydrogen gas produced, use the molar mass of hydrogen. The molar mass of hydrogen gas ({H}_2) is approximately 2.02 g/mol. Calculate the moles using the formula:\[moles\ of\ H_2 = \frac{mass\ of\ H_2}{molar\ mass\ of\ H_2} = \frac{0.1152\ g}{2.02\ g/mol} \ \approx 0.057\ moles\]
02

Use Stoichiometry to Find Moles of Iron

Using the balanced chemical equation, we know that 3 moles of hydrogen gas are produced for every 2 moles of iron that react. Use this to find the moles of iron:\[\frac{2\ moles\ Fe}{3\ moles\ H_2} = \frac{x\ moles\ Fe}{0.057\ moles\ H_2}\]Solving for \(x\) yields:\[x = \frac{2}{3} \times 0.057 \x \approx 0.038\ moles\ Fe\]
03

Calculate Mass of Iron in the Alloy

Now calculate the mass of iron using moles and its atomic mass. The atomic mass of iron (Fe) is approximately 55.85 g/mol. Thus, the mass of iron is:\[mass\ of\ Fe = moles\ of\ Fe \times atomic\ mass\ of\ Fe = 0.038\ moles \times 55.85\ g/mol \mass\ of\ Fe \approx 2.12\ g\]
04

Calculate Percentage of Iron in the Alloy

To determine the percentage of iron in the alloy, use the formula for percentage by mass:\[percentage\ of\ Fe = \left(\frac{mass\ of\ Fe}{total\ mass\ of\ alloy}\right) \times 100 = \left(\frac{2.12\ g}{2.358\ g}\right) \times 100 \percentage\ of\ Fe \approx 89.92\%\]

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Key Concepts

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

Moles of hydrogen
The term "moles of hydrogen" refers to the quantity of hydrogen gas, a common element involved in chemical reactions. Hydrogen gas is made up of two hydrogen atoms forming \( ext{H}_2 \). In chemical terms, a mole is a unit that measures the amount of substance, equivalent to Avogadro鈥檚 number, which is approximately \(6.022 imes 10^{23}\) entities, such as atoms or molecules.
To determine the moles of hydrogen produced in a chemical reaction, we use the ratio of the mass of hydrogen produced to its molar mass.
In the given exercise, we calculate the moles of hydrogen gas produced by dividing the mass of hydrogen (0.1152 g) by its molar mass (2.02 g/mol). This straightforward division gives us approximately 0.057 moles of hydrogen, indicating the amount of hydrogen gas generated in the reaction.
Moles of iron
Moles of iron refers to the amount of iron participating in the chemical reaction. In our exercise, the relationship between hydrogen and iron from the balanced equation is crucial.
According to the balanced chemical equation for the reaction, 2 moles of iron ( 2 Fe) are needed to produce 3 moles of hydrogen gas ( 3 H_2). Thus, this ratio allows us to find the number of moles of iron that reacted, given the number of moles of hydrogen produced.
We previously calculated the moles of hydrogen as 0.057, and using this information along with the stoichiometric ratio ( 2 moles of Fe / 3 moles of H_2), we can determine that approximately 0.038 moles of iron were involved in the reaction. This calculation helps us link the amount of hydrogen produced back to the initial amount of iron that reacted with sulfuric acid.
Percentage composition
Percentage composition refers to the relative amount of each element within a compound. It enables us to understand the proportion of a particular element in a mixture, critical for performing analyses and comparisons in chemical stoichiometry.
In the exercise, we need to determine the percentage of iron in the iron-carbon alloy by calculating what percentage of the total mass is made up by iron.
The formula for percentage composition by mass is: \(\text{percentage of Fe} = \left(\frac{\text{mass of Fe}}{\text{total mass of alloy}}\right) \times 100 \).
By substituting the mass of iron (2.12 g) and the total mass of the sample (2.358 g) into this formula, we find that the percentage of iron in the alloy is approximately 89.92\%. This important calculation allows us to quantify exactly how much of the sample is iron, providing insight into the alloy's composition.
Mass of iron
The mass of iron is a straightforward but often critical calculation in stoichiometry. Once we know the moles of iron involved in a reaction, we can easily find its mass using iron's atomic mass.
Iron has an atomic mass of approximately 55.85 g/mol. By multiplying this atomic mass by the moles of iron, which we previously calculated to be 0.038, we determine the mass of iron in the alloy.
It's calculated as follows: \[\text{mass of Fe} = 0.038\, \text{moles} \times 55.85\, \text{g/mol} \approx 2.12\, \text{g} \].
This calculation tells us how much of the alloy's total mass is made up of iron atoms, which is critical when determining the purity and quality of metal alloys.

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Most popular questions from this chapter

Describe in words how you would do each of the following preparations. Then give the molecular equation for each preparation. a. \(\mathrm{CuCl}_{2}(s)\) from \(\mathrm{CuSO}_{4}(s)\) b. \(\mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(s)\) from \(\mathrm{CaCO}_{3}(s)\) c. \(\mathrm{NaNO}_{3}(s)\) from \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) d. \(\mathrm{MgCl}_{2}(s)\) from \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

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An alloy of aluminum and magnesium was treated with sodium hydroxide solution, in which only aluminum reacts. \(2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow{2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g)}\) If a sample of alloy weighing \(1.118\) g gave \(0.1068 \mathrm{~g}\) of hydrogen, what is the percentage of aluminum in the alloy?

Classify each of the following reactions as a combination reaction, decomposition reaction, displacement reaction, or combustion reaction. a. When solid calcium oxide, \(\mathrm{CaO}\), is exposed to gaseous sulfur trioxide, \(\mathrm{SO}_{3}\), solid calcium sulfate, \(\mathrm{CaSO}_{4}\), is formed. b. Calcium metal (solid) reacts with water to produce a solution of calcium hydroxide, \(\mathrm{Ca}(\mathrm{OH})_{2}\), and hydrogen gas. c. When solid sodium hydrogen sulfite, \(\mathrm{NaHSO}_{3}\), is heated, solid sodium sulfite, \(\mathrm{Na}_{2} \mathrm{SO}_{3}\), sulfur dioxide gas, \(\mathrm{SO}_{2}\), and water vapor are formed. d. Magnesium reacts with bromine to give magnesium bromide, \(\mathrm{MgBr}_{2}\).
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