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Sulfuric acid can be produced from a sulfide ore such as iron pyrite by the following sequence of reactions: $$\begin{aligned} 4 \mathrm{FeS}_{2}(\mathrm{s})+11 \mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+8 \mathrm{SO}_{2}(\mathrm{g}) \\ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{g}) \\ \mathrm{SO}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) & \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \end{aligned}$$ Starting with \(525 \mathrm{kg}\) of \(\mathrm{FeS}_{2}\) (and an excess of other reactants), what mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) can be prepared?

Short Answer

Expert verified
857.3 kg of H2SO4 can be prepared.

Step by step solution

01

Convert Mass of FeS2 to Moles

First, we need the molar mass of FeS2. The molar mass of Fe is 55.85 g/mol, and S is 32.07 g/mol. From this, the molar mass of FeS2 is calculated as: \[(1 \times 55.85) + (2 \times 32.07) = 119.99 \text{ g/mol}.\] Now, convert 525 kg of FeS2 to grams (since the molar mass is in g/mol): \[525 \text{ kg} = 525,000 \text{ g}.\] Calculate the moles of FeS2: \[\frac{525,000 \text{ g}}{119.99 \text{ g/mol}} \approx 4370 \text{ mol}.\]
02

Reaction Stoichiometry of FeS2

According to the first reaction: \[4 \mathrm{FeS}_2 \rightarrow 8 \mathrm{SO}_2.\] From 4 moles of FeS2, 8 moles of SO2 are produced. Thus, each mole of FeS2 produces 2 moles of SO2. Calculate moles of SO2 formed: \[4370 \text{ mol FeS2} \times 2 = 8740 \text{ mol SO2}.\]
03

Reaction Stoichiometry of SO2 to SO3

In the second reaction: \[2 \mathrm{SO}_2 + \mathrm{O}_2 \rightarrow 2 \mathrm{SO}_3.\] 2 moles of SO2 yield 2 moles of SO3, meaning a 1:1 conversion. Thus, 8740 moles of SO2 produce 8740 moles of SO3.
04

Convert Moles of SO3 to Moles of H2SO4

The third reaction is: \[\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_2\mathrm{SO}_4.\] Each mole of SO3 reacts to form one mole of H2SO4. Therefore, 8740 moles of SO3 result in 8740 moles of H2SO4.
05

Convert Moles of H2SO4 to Mass

The molar mass of H2SO4 is calculated as: \[(2 \times 1.01) + (32.07) + (4 \times 16.00) = 98.09 \text{ g/mol}.\] Calculate the mass of H2SO4 produced: \[8740 \text{ mol} \times 98.09 \text{ g/mol} = 857,308.6 \text{ g}.\] Convert this mass to kilograms to find: \[857,308.6 \text{ g} \approx 857.3 \text{ kg}.\]

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Key Concepts

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

Stoichiometry
Stoichiometry is the heart of chemical calculations. It involves using balanced chemical equations to determine the relationships between reactants and products. In the sulfuric acid production from iron pyrite, stoichiometry helps us calculate how much of each substance is consumed or produced.
  • We use the coefficients in the balanced equation to find the molar ratios between different substances.
  • For example, in the conversion of iron pyrite (\( \mathrm{FeS}_2 \)) to sulfur dioxide (\( \mathrm{SO}_2 \)), the ratio is 4:8, meaning 4 moles of \( \mathrm{FeS}_2 \) produces 8 moles of \( \mathrm{SO}_2 \)
This quantitative relationship allows us to move from moles of one compound to moles of another, crucial for determining yields in chemical reactions.
Chemical Reactions
Chemical reactions are processes where substances transform into different substances. In the production of sulfuric acid, we have a series of reactions starting from iron pyrite.
  • The first reaction involves the burning of iron pyrite in oxygen to form iron oxide and sulfur dioxide.
  • The sulfur dioxide further reacts with oxygen to give sulfur trioxide.
  • Finally, sulfur trioxide reacts with water to form sulfuric acid.
Each step is crucial and depends on the previous step's products. Understanding these reactions helps us trace the path and transformation of materials in sulfuric acid production.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance and is essential in converting between mass and moles. Calculating the molar mass of compounds like iron pyrite and sulfuric acid enables us to work out the amount involved in reactions.
  • The molar mass of iron pyrite (\( \mathrm{FeS}_2 \)) is calculated by adding the atomic masses of iron and sulfur.
  • Similarly, the molar mass of sulfuric acid (\( \mathrm{H}_2\mathrm{SO}_4 \)) is found by summing up the masses of hydrogen, sulfur, and oxygen.
Knowing these values allows us to easily convert grams to moles and vice versa, enabling accurate stoichiometric calculations.
Iron Pyrite
Iron pyrite, often nicknamed "fool's gold," is a common sulfide mineral. It's chemically represented as \( \mathrm{FeS}_2 \). In the context of sulfuric acid production, iron pyrite acts as a starting material.
  • When heated in the presence of oxygen, it decomposes to produce sulfur dioxide, a critical intermediate in the process.
  • The transformation from \( \mathrm{FeS}_2 \) is what sets the stage for the subsequent steps leading to sulfuric acid.
Its importance lies in its ability to release sulfur, a key sulfide, necessary for the production of sulfuric acid. This makes iron pyrite a significant raw material in the chemical industry.

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Most popular questions from this chapter

Sulfur dioxide, a pollutant produced by burning coal and oil in power plants, can be removed by reaction with calcium carbonate. \(2 \mathrm{SO}_{2}(\mathrm{g})+2 \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CaSO}_{4}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g})\) (a) What mass of \(\mathrm{CaCO}_{3}\) is required to remove \(155 \mathrm{g}\) of \(\mathrm{SO}_{2} ?\) (b) What mass of \(\mathrm{CaSO}_{4}\) is formed when \(155 \mathrm{g}\) of \(\mathrm{SO}_{2}\) is consumed completely?

An unknown metal reacts with oxygen to give the metal oxide, MO \(_{2} .\) Identify the metal based on the following information: Mass of metal \(=0.356 \mathrm{g}\) Mass of sample after converting metal completely to oxide \(=0.452 \mathrm{g}\)

Aluminum chloride, AlCl_, is made by treating scrap eluminum with chlorine. $$ 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{s}) $$ If you begin with \(2.70 \mathrm{g}\) of \(\mathrm{Al}\) and \(4.05 \mathrm{g}\) of \(\mathrm{Cl}_{2}\) (a) Which reactant is limiting? (b) What mass of AlCl \(_{3}\) can be produced? (c) What mass of the excess reactant remains when the reaction is completed?

Balance the following equations: (a) \(\operatorname{Cr}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\) (b) \(\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)

Sodium azide, the explosive chemical used in automobile airbags, is made by the following reaction: $$ \mathrm{NaNO}_{3}+3 \mathrm{NaNH}_{2} \longrightarrow \mathrm{NaN}_{3}+3 \mathrm{NaOH}+\mathrm{NH}_{3} $$ If you combine \(15.0 \mathrm{g}\) of \(\mathrm{NaNO}_{3}(85.0 \mathrm{g} / \mathrm{mol})\) with \(15.0 \mathrm{g}\) of \(\mathrm{NaNH}_{2},\) what mass of \(\mathrm{NaN}_{3}\) is produced?

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