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Chapter 5: Problem 144

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when 120. g of methane is reacted with an equal mass of sulfur.

Short Answer

Expert verified
The balanced chemical equation for the reaction of methane and sulfur is \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\). When 120 g of methane is reacted with an equal mass of sulfur, the theoretical yield of carbon disulfide is 71.2 g.

Step by step solution

01

Write the unbalanced chemical equation

The unbalanced chemical equation is: \(CH_4 + S \rightarrow CS_2 + H_2S\)
02

Balance the chemical equation

To balance the chemical equation, we need to have equal numbers of each element on both sides of the equation. Only the hydrogen atoms are not balanced now, so we should add a coefficient of 2 in front of H2S. The balanced chemical equation is: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\) b. Theoretical yield of carbon disulfide
03

Calculate the number of moles of methane and sulfur

Using the molar mass of methane (16.04 g/mol) and sulfur (32.07 g/mol), calculate the number of moles of both reactants. Moles of methane: \[\frac{120 \, g}{16.04 \, g/mol} = 7.48 \, mol\] Moles of sulfur: \[\frac{120 \, g}{32.07 \, g/mol} = 3.74 \, mol\]
04

Identify the limiting reactant

Compare the mole ratios of methane and sulfur in the balanced chemical equation with the mole ratios of the reactants. Ratio of methane: \[\frac{7.48 \, mol}{1} = 7.48\] Ratio of sulfur: \[\frac{3.74 \, mol}{4} = 0.935\] Since the ratio of sulfur is smaller than the ratio of methane, sulfur is the limiting reactant.
05

Calculate the theoretical yield of carbon disulfide

Use the mole ratio of sulfur to carbon disulfide from the balanced chemical equation and the molar mass of CS2 (76.13 g/mol) to calculate the theoretical yield. Moles of CS2: \[\text{Moles of S} \times \frac{\text{Moles of CS2}}{\text{Moles of S}} = 3.74 \, mol \times \frac{1 \, mol \, CS2}{4 \, mol \, S} = 0.935 \, mol \, CS2\] Theoretical yield of CS2: \[0.935 \, mol \times 76.13 \, g/mol = 71.2 \, g\] Therefore, the theoretical yield of carbon disulfide is 71.2 g.

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Most popular questions from this chapter

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react \(52.9 \mathrm{g}\) of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide \(\left(\mathbf{P}_{4} \mathbf{O}_{10}\right)\) could be produced? $$\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad \text { (unbalanced) }$$

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

Consider the following data for three binary compounds of hydrogen and nitrogen: $$\begin{array}{|lcc|}\hline & \% \text { H (by Mass) } & \% \text { N (by Mass) } \\\\\hline \text { I } & 17.75 & 82.25 \\\\\text { II } & 12.58 & 87.42 \\\\\text { III } & 2.34 & 97.66 \\\\\hline\end{array}$$ When 1.00 L of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$\begin{array}{|lcc|}\hline & \mathrm{H}_{2}(\mathrm{L}) & \mathrm{N}_{2}(\mathrm{L}) \\\\\hline \\\\\mathrm{I} & 1.50 & 0.50 \\\\\mathrm{II} & 2.00 & 1.00 \\\\\mathrm{III} & 0.50 & 1.50 \\\\\hline\end{array}$$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

A \(9.780-\mathrm{g}\) gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Complete combustion to form carbon dioxide and water requires 1.120 mole of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \operatorname{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)
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