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Chapter 9: Problem 82

One step in the commercial production of sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), involves the conversion of sulfur dioxide, \(\mathrm{SO}_{2},\) into sulfur trioxide, \(\mathrm{SO}_{3}\). $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) $$ If \(150 \mathrm{~kg}\) of \(\mathrm{SO}_{2}\) reacts completely, what mass of \(\mathrm{SO}_{3}\) should result?

Short Answer

Expert verified
When 150 kg of SO鈧� reacts completely, 187.5 kg of SO鈧� should be produced.

Step by step solution

01

Convert mass of SO鈧� to moles

The molar mass of sulfur dioxide (SO鈧�) is 32.07 (S) + 2 脳 16.00 (O) = 64.07 g/mol. We are given that 150 kg of SO鈧� reacts completely. To convert this mass to moles, we can use the formula: moles = mass / molar mass So in this case, moles of SO鈧� = \( \frac{150,000\,\text{g}}{64.07\,\text{g/mol}} \approx 2341.5\,\text{moles} \)
02

Use stoichiometry to find moles of SO鈧�

From the balanced chemical equation, we can see that 2 moles of SO鈧� react with 1 mole of O鈧� to produce 2 moles of SO鈧�. The stoichiometric ratio of SO鈧� to SO鈧� is 2:2, which simplifies to 1:1. Therefore, for each mole of SO鈧� consumed, one mole of SO鈧� is produced. So, the moles of SO鈧� produced are equal to the moles of SO鈧� consumed: moles of SO鈧� = 2341.5 moles
03

Convert moles of SO鈧� to mass

The molar mass of sulfur trioxide (SO鈧�) is 32.07 (S) + 3 脳 16.00 (O) = 80.07 g/mol. Now we can convert the moles of SO鈧� back to mass using the formula: mass = moles 脳 molar mass mass of SO鈧� = 2341.5 moles 脳 80.07 g/mol 鈮� 187531.5 g = 187.5 kg (rounded to one decimal place) So, when 150 kg of SO鈧� reacts completely, 187.5 kg of SO鈧� should be produced.

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Key Concepts

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

Understanding Molar Mass
Molar mass is a central concept in stoichiometry, as it connects the mass of a substance to the amount in moles. It is defined as the mass of one mole of a given substance in grams. To find the molar mass, sum up the atomic masses of all atoms in a chemical formula.

For example, the molar mass of sulfur dioxide (\(\text{SO}_2\)) is calculated by adding:
  • The atomic mass of one sulfur atom: 32.07 g/mol
  • Twice the atomic mass of oxygen (since there are two oxygen atoms): 2 \(\times\) 16.00 g/mol
Thus, the molar mass of \(\text{SO}_2\) is 64.07 g/mol.

Similarly, for sulfur trioxide (\(\text{SO}_3\)), the molar mass would be 32.07 g/mol for sulfur plus three times 16.00 g/mol for the oxygen atoms, equalling 80.07 g/mol. Knowing these values allows us to convert between mass and moles, enabling us to predict the outcomes of chemical reactions.
Basics of Chemical Reactions
Chemical reactions involve the transformation of reactants into products. They are represented by balanced chemical equations that show the relationship between the molecules involved in the reaction. The balanced equation indicates the proportion or ratio in which substances react and form products.

In the equation \(2\, \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\, \text{SO}_3(g)\), we see that sulfur dioxide reacts with oxygen to form sulfur trioxide. Here, the coefficients show that two moles of \(\text{SO}_2\) react with one mole of \(\text{O}_2\) to produce two moles of \(\text{SO}_3\).

This equation is balanced because the number of each type of atom is the same on both sides of the equation. Balanced equations allow chemists to calculate the necessary amounts of reactants and predict the amount of product formed.
The Process of Mole Conversion
Mole conversion is a technique used to transition between units of mass and moles. It is essential in stoichiometry because most chemical reactions are described in terms of moles.

When converting from mass to moles, use the formula:
  • Moles = Mass / Molar Mass
For example, if you have 150 kg of \(\text{SO}_2\), convert it into grams (150,000 g) and then find the number of moles by dividing by its molar mass (64.07 g/mol), resulting in about 2341.5 moles of \(\text{SO}_2\).

Once in moles, stoichiometric ratios from a balanced equation allow conversion between substances. Since 1 mole of \(\text{SO}_2\) produces 1 mole of \(\text{SO}_3\), the moles of \(\text{SO}_3\) will also be 2341.5. To find the resulting mass of \(\text{SO}_3\), you multiply the moles by the molar mass of \(\text{SO}_3\) (80.07 g/mol), leading to a mass of 187.5 kg. This process helps accurately evaluate the quantities involved in chemical reactions.

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Most popular questions from this chapter

If common sugars are heated too strongly, they char as they decompose into carbon and water vapor. For example, if sucrose (table sugar) is heated, the reaction is $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12 \mathrm{C}(s)+11 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of carbon is produced if 1.19 g of sucrose decomposes completely?

The traditional method of analysis for the amount of chloride ion prescnt in a sample is to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample. Suppose a \(1.054-g\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples. a. 2.17 moles of germanium, Ge b. \(4.24 \mathrm{mmol}\) of lead(II) chloride ( \(1 \mathrm{mmol}=1 / 1000 \mathrm{~mol}\) ) c. 0.0971 mole of ammonia, \(\mathrm{NH}_{3}\) d. \(4.26 \times 10^{3}\) moles of hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\) e. 1.71 moles of iodine monochloride, \(1 \mathrm{C}=\)

Using the average atomic masses given inside the front cover of the text, calculate the mass in grams of cach of the following samples. a. 5.0 moles of nitric acid b. 0.000305 mole of mercury c. \(2.31 \times 10^{-5}\) mole of potassium chromate d. 10.5 moles of aluminum chloride e. \(4.9 \times 10^{4}\) moles of sulfur hexafluoride f. 125 moles of ammonia g. 0.01205 mole of sodium peroxide

For cach of the following balanced chemical cquations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion. a. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathrm{CH}_{4}(g)+4 \mathrm{~S}(s) \rightarrow \mathrm{CS}_{2}(l)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathbf{P C l}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HCl}(a q)\) d. \(\mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{NaHCO}_{3}(s)\)
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