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Chapter 10: Problem 28

In the contact process, sulfur dioxide and oxygen gas react to form sulfur trioxide as follows: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ At a certain temperature and pressure, \(50 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) reacts with \(25 \mathrm{~L}\) of \(\mathrm{O}_{2}\). If all the \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are consumed, what volume of \(\mathrm{SO}_{3}\), at the same temperature and pressure, will be produced?

Short Answer

Expert verified
When \(50 L\) of \(SO_2\) and \(25 L\) of \(O_2\) react completely at the same temperature and pressure, \(50 L\) of \(SO_3\) will be produced, according to the stoichiometry of the balanced chemical equation.

Step by step solution

01

Determine the stoichiometric ratio

In the given balanced chemical equation, the stoichiometric ratio between the reactants and the product is \(2 \mathrm{SO}_{2} : 1 \mathrm{O}_{2} : 2 \mathrm{SO}_{3}\). This means for every two moles of \(SO_2\), one mole of \(O_2\) is required to produce two moles of \(SO_3\). We can apply this ratio to volumes as well, since they are directly proportional to moles at the same temperature and pressure.
02

Identify the limiting reactant

Now, we find the limiting reactant: the reactant that gets consumed first and hence determines the amount of product formed. We have \(50 L\) of \(SO_2\) and \(25 L\) of \(O_2\). According to the stoichiometric ratio, we need two volumes of \(SO_2\) for one volume of \(O_2\). So, for \(25 L\) of \(O_2\), we would need \(2 \times 25 L = 50 L\) of \(SO_2\). Since we have exactly \(50 L\) of \(SO_2\), it is the limiting reactant and both reactants will be completely consumed.
03

Calculate the volume of SO3 formed

Once we know that the limiting reactant is \(SO_2\), we can determine the volume of \(SO_3\) produced. According to the stoichiometric ratio, for every 2 volumes of \(SO_2\), 2 volumes of \(SO_3\) are produced. Since there are \(50 L\) of \(SO_2\), we can calculate the volume of \(SO_3\) formed as follows: Volume of \(SO_3\) = (Volume of \(SO_2\))/(Volume ratio of \(SO_2\)) × (Volume ratio of \(SO_3\)) = \(\frac{50 L}{2} \times 2 = 50 L\) Hence, \(50 L\) of \(SO_3\) will be produced at the same temperature and pressure when all the \(SO_2\) and \(O_2\) are consumed.

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Most popular questions from this chapter

Table 10.3 shows that the van der Waals \(b\) parameter has units of $\mathrm{L} / \mathrm{mol}$. This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is $(4 / 3) \pi r^{3}$.

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Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If $1.56 \mathrm{~g}\( of cyclopropane has a volume of \)1.00 \mathrm{~L}$ at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

(a) What are the mole fractions of each component in a mixture of $15.08 \mathrm{~g}\( of \)\mathrm{O}_{2}, 8.17 \mathrm{~g}\( of \)\mathrm{N}_{2},$ and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)
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