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Chapter 3: Problem 85

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(S_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\), how many grams of \(S_{8}\) would be produced, assuming \(98 \%\) yield?

Short Answer

Expert verified
The actual mass of S8 produced in this reaction, assuming a 98% yield, is \(27.6 \, grams\).

Step by step solution

01

Convert masses to moles

First, we need to convert the masses of H2S and O2 to moles. We use their respective molar masses for this conversion. Molar mass of H2S: \(1 (H) * 2 + 32 (S) = 2 + 32 = 34 g/mol\) Molar mass of O2: \(16 (O) * 2 = 32 g/mol\) Now, we can convert the masses of H2S and O2 to moles: Moles of H2S = \(\dfrac{30.0 g}{34 g/mol} = 0.882 moles\) Moles of O2 = \(\dfrac{50.0 g}{32 g/mol} = 1.563 moles\)
02

Identify the limiting reactant

Next, we need to identify the limiting reactant. For this, we will use the mole ratios provided by the balanced equation: \(8 \, H_2S + 4 \, O_2 \rightarrow S_8 + 8 \, H_2O\) Divide moles of each reactant by their respective stoichiometric coefficients: Moles ratio of H2S: \(\dfrac{0.882}{8} = 0.110\) Moles ratio of O2: \(\dfrac{1.563}{4} = 0.391\) Since H2S has the lowest mole ratio, it will be the limiting reactant.
03

Calculate the theoretical yield of S8

Now, we need to calculate the theoretical yield of S8 in moles, using the balanced equation. In this reaction, 8 moles of H2S produce 1 mole of S8. Moles of S8 = moles of H₂S * \(\dfrac{1 \, mole \, S_{8}}{8 \, moles \, H_{2}S}\) = \(0.882 * \dfrac{1}{8} = 0.110\) moles Now, let's calculate the theoretical yield in grams. The molar mass of S8 is: \(32 (S) * 8 = 256 g/mol\). Theoretical yield of S8 = \(0.110 \, moles \times 256 \dfrac{g}{mol}\) = \(28.16 g\)
04

Apply 98% yield to find actual mass of S8 produced

Finally, we need to apply the given 98% yield to find the actual mass of S8 produced: Actual mass of S8 = theoretical yield of S8 * \% yield = \(28.16 g * 0.98 = 27.6 g\) Therefore, the actual mass of S8 produced in this reaction is 27.6 grams.

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Most popular questions from this chapter

The thermite reaction, $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{Fe} $$ produces so much heat that the Fe product melts. This reaction is used industrially to weld metal parts under water, where a torch cannot be employed. It is also a favorite chemical demonstration in the lecture hall (on a small scale). (a) Balance the chemical equation for the thermite reaction, and include the proper states of matter. (b) Calculate how many grams of aluminum are needed to completely react with \(500.0 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in this reaction. (c) This reaction produces \(852 \mathrm{~kJ}\) of heat per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacted. How many grams of $\mathrm{Fe}_{2} \mathrm{O}_{3}\( are needed to produce \)1.00 \times 10^{4} \mathrm{~kJ}$ of heat? (d) If you performed the reverse reaction- aluminum oxide plus iron makes iron oxide plus aluminum-would that reaction have heat as a reactant or a product?

Calculate the percentage by mass of the indicated element in the following compounds: (a) hydrogen in methane, \(\mathrm{CH}_{4}\), the major hydrocarbon in natural gas; \((\mathbf{b})\) oxygen in vitamin $\mathrm{E}, \mathrm{C}_{29} \mathrm{H}_{50} \mathrm{O}_{2} ;$ (c) sulphur in magnesium sulphate, \(\mathrm{MgSO}_{4}\), a substance used as a drying agent; \((\mathbf{d})\) nitrogen in epinephrine, \(\mathrm{C}_{9} \mathrm{H}_{13} \mathrm{NO}_{3},\) also known as adrenalin, a hormone that is important for the fightor-flight response; (e) oxygen in the insect pheromone sulcatol, $\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O} ;\( (f) carbon in sucrose, \)\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11},$ the compound that is responsible for the sweet taste of table sugar.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) $\mathrm{C}_{7} \mathrm{H}_{16}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ (b) $\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)$ (c) $\mathrm{Zn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(I)$ (d) $\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Ag}(s)+\mathrm{O}_{2}(g)$

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate \(\left(\mathrm{NaHCO}_{3}\right)\) and citric acid $\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)$ $$ \begin{aligned} 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) & \longrightarrow \\ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) \end{aligned} $$ In a certain experiment \(1.00 \mathrm{~g}\) of sodium bicarbonate and $1.00 \mathrm{~g}$ of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed?

(a) What is the mass, in grams, of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate? (b) How many moles of chloride ions are in \(0.2550 \mathrm{~g}\) of aluminum chloride? (c) What is the mass, in grams, of \(7.70 \times 10^{20}\) molecules of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\) (d) What is the molar mass of cholesterol if \(0.00105 \mathrm{~mol}\) has a mass of \(0.406 \mathrm{~g}\) ?
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