/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Disulfur dichloride, \(\mathrm{S... [FREE SOLUTION] | 91Ó°ÊÓ

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Disulfur dichloride, \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), is used to vulcanize rubber. It can be made by treating molten sulfur with gaseous chlorine: $$ \mathrm{S}_{\mathrm{s}}(\ell)+4 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}(\ell) $$ Starting with a mixture of \(32.0 \mathrm{g}\) of sulfur and \(71.0 \mathrm{g}\) of \(\mathrm{Cl}_{2}\) (a) Which is the limiting reactant? (b) What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2} ?\) (c) What mass of the excess reactant remains when the reaction is completed?

Short Answer

Expert verified
(a) \( \mathrm{Cl}_2 \) is the limiting reactant. (b) 33.76 g of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is the theoretical yield. (c) 24.05 g of sulfur remains.

Step by step solution

01

- Calculate Moles of Reactants

First, find the molar mass of sulfur, \( \mathrm{S} \), which is \( 32.07\, \mathrm{g/mol} \). Then, calculate the moles of sulfur: \( \frac{32.0\, \mathrm{g}}{32.07\, \mathrm{g/mol}} = 1.00\, \mathrm{mol} \).Next, find the molar mass of chlorine, \( \mathrm{Cl}_2 \), which is \( 70.9\, \mathrm{g/mol} \). Calculate the moles of chlorine: \( \frac{71.0\, \mathrm{g}}{70.9\, \mathrm{g/mol}} = 1.00\, \mathrm{mol} \).
02

- Determine Limiting Reactant

The balanced equation is: \[ \mathrm{S}_{\mathrm{s}}(\ell) + 4 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}(\ell) \]According to the equation, \( 1 \) mole of sulfur reacts with \( 4 \) moles of \( \mathrm{Cl}_2 \). Here, \( 1.00 \) mole of sulfur would require \( 4.00 \) moles of \( \mathrm{Cl}_2 \), but we only have \( 1.00 \) mole of \( \mathrm{Cl}_2 \). Thus, \( \mathrm{Cl}_2 \) is the limiting reactant.
03

- Calculate Theoretical Yield of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \)

Since \( \mathrm{Cl}_2 \) is the limiting reactant and \( \frac{1}{4} \) mole of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is produced per mole of \( \mathrm{Cl}_2 \), the theoretical yield is:\[ 1.00\, \text{mol} \times \frac{1}{4} = 0.25\, \text{mol of } \mathrm{S}_{2} \mathrm{Cl}_{2} \]The molar mass of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is \( 135.04\, \mathrm{g/mol} \).Calculate the mass: \( 0.25\, \text{mol} \times 135.04\, \mathrm{g/mol} = 33.76\, \mathrm{g} \).
04

- Calculate Mass of Excess Reactant Remaining

Initially, \( 1.00 \) mole of \( \mathrm{Cl}_2 \) reacts completely, requiring \( \frac{1.00}{4} = 0.25 \) moles of sulfur.The initial moles of sulfur are \( 1.00 \) mole, so the moles of sulfur left are \( 1.00 - 0.25 = 0.75 \) moles.Find the mass of the leftover sulfur: \( 0.75\, \text{mol} \times 32.07\, \mathrm{g/mol} = 24.05\, \mathrm{g} \).

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Key Concepts

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

Mole Calculation
Mole calculation is a fundamental aspect of chemistry that helps in determining how much of each substance participates in a chemical reaction. When calculating moles, you need to know the molar mass of each reactant, which you can find on the periodic table. The molar mass provides the link between grams of a substance and the amount in moles. In our exercise example, we calculate the moles of sulfur and chlorine using their respective molar masses:
  • For sulfur (\( \mathrm{S} \)), with a molar mass of \( 32.07\, \mathrm{g/mol} \)
  • For chlorine (\( \mathrm{Cl}_2 \)), with a molar mass of \( 70.9\, \mathrm{g/mol} \)
By dividing the mass of each reactant by its molar mass, you calculate the moles involved in the reaction.To further understand and accurately perform such calculations, it's essential to be comfortable with converting between grams and moles. This step is crucial for determining how much of each chemical is needed and what remains after the reaction. Always check your calculations to ensure accuracy before solving further steps of a chemical reaction.
Theoretical Yield
The theoretical yield refers to the maximum amount of product that can be produced in a chemical reaction based on the stoichiometry and the limiting reactant. It is calculated by using the balanced chemical equation to determine the proportion of reactants involved. In our case, the limiting reactant is chlorine, \( \mathrm{Cl}_2 \), since we have less than what the stoichiometry requires.Once the limiting reactant is identified, the theoretical yield is computed by relating moles of the limiting reactant to moles of the desired product. According to the reaction equation, every 4 moles of \( \mathrm{Cl}_2 \) produces 1 mole of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).Given we have 1 mole of \( \mathrm{Cl}_2 \), it results in \( 0.25 \) moles of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).Finally, to find the mass of the theoretical yield, you multiply the moles of product by its molar mass, \( 135.04\, \mathrm{g/mol} \), which yields \( 33.76\, \mathrm{g} \) in this example. This process highlights how understanding stoichiometry and limiting reactants can predict the potential outcome of reactions before they are carried out in the lab.
Stoichiometry
Stoichiometry is the heart of balancing chemical reactions and allows chemists to predict how much of each substance will react, be produced, or remain. It is based on the balanced chemical equation, which uses relative quantities of reactants and products in moles.A stoichiometrically balanced equation ensures that the number of atoms for each element is equal on both sides of the equation. In this exercise, the equation shows that 1 mole of sulfur reacts with 4 moles of chlorine to produce 4 moles of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).When considering which reactant limits the reaction, stoichiometry is essential because it relies on these mole ratios. Identifying the limiting reactant—chlorine in this situation—helps determine how much \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) can form and how much sulfur will be in excess after the reaction. Stoichiometry also helps calculate remaining quantities once the reaction is complete, allowing chemists to optimize processes, save resources, and minimize waste.

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Most popular questions from this chapter

A Chromium(III) chloride forms many compounds with ammonia. To find the formula of one of these compounds, you titrate the \(\mathrm{NH}_{3}\) in the compound with standardized acid. \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}} \mathrm{Cl}_{3}(\mathrm{aq})+\times \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ x \mathrm{NH}_{4}+(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq})+(x+3) \mathrm{Cl}^{-}(\mathrm{aq}) $$ Assume that \(24.26 \mathrm{mL}\) of \(1.500 \mathrm{M} \mathrm{HCl}\) is used to tirate \(1.580 \mathrm{g}\) of \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{3} .\) What is the value of \(x ?\)

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

A Copper metal can be prepared by roasting copper ore, which can contain cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right)\) and copper (11) sulfide. $$ \begin{aligned} \mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{CuS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) & \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \end{aligned} $$ Suppose an ore sample contains \(11.0 \%\) impurity in addition to a mixture of CuS and Cu \(_{2} \mathrm{S}\). Heating \(100.0 \mathrm{g}\) of the mixture produces \(75.4 \mathrm{g}\) of copper metal with a purity of \(89.5 \% .\) What is the weight percent of CuS in the ore? The weight percent of \(\mathrm{Cu}_{2} \mathrm{S} ?\)

A Boron forms a series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\) $$ \mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\text { excess } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If \(0.148 \mathrm{g}\) of one of these compounds gives \(0.422 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2},\) what is its empirical formula?

If you dilute \(25.0 \mathrm{mL}\) of \(1.50 \mathrm{M}\) hydrochloric acid to \(500 . \mathrm{mL},\) what is the molar concentration of the dilute acid?

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