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The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Short Answer

Expert verified
The theoretical mass of As鈧侷鈧 that can be prepared is 670.20 g, and the actual mass of As鈧侷鈧 isolated with a percent yield of 75.6% is 506.51 g.

Step by step solution

01

Calculate the mass of arsenic metal

To calculate the mass of arsenic metal, we need to use the density formula as follows: mass = density 脳 volume. The volume of the arsenic cubic block can be calculated as side 脳 side 脳 side. Density = 5.72 g/cm鲁 Side = 3.00 cm Volume = 3.00 cm 脳 3.00 cm 脳 3.00 cm = 27 cm鲁 Now, we can calculate the mass of arsenic: Mass of arsenic = density 脳 volume = 5.72 g/cm鲁 脳 27 cm鲁 = 154.44 g
02

Calculate the moles of arsenic and arsenic triiodide

To determine the moles of arsenic and arsenic triiodide, we will use the following relationship: moles = mass / molar mass Molar mass of arsenic (As) = 74.92 g/mol Moles of arsenic = mass of arsenic / molar mass of arsenic = 154.44 g / 74.92 g/mol = 2.061 mol We are given the number of molecules of arsenic triiodide, so we need to convert it to moles. We will use Avogadro's number, which is 6.022 脳 10虏鲁 molecules/mol: Moles of arsenic triiodide = (1.01 脳 10虏鈦 molecules) / (6.022 脳 10虏鲁 molecules/mol) = 1.678 mol
03

Determine the limiting reactant

To find out the limiting reactant, we need to compare the molar ratio of arsenic to arsenic triiodide with the stoichiometric ratio. The balanced chemical equation for the reaction is: 2 As + 2 AsI鈧 鈫 As鈧侷鈧 Stoichiometric ratio of As to AsI鈧 = 1 : 1 Molar ratio of As to AsI鈧 = 2.061 mol : 1.678 mol = 1 : 0.815 Since the molar ratio of As to AsI鈧 is greater than the stoichiometric ratio, arsenic triiodide is the limiting reactant.
04

Calculate the theoretical mass of As鈧侷鈧 that can be prepared

As AsI鈧 is the limiting reactant, we can use the stoichiometry of the reaction to find the theoretical amount of As鈧侷鈧 that can be produced: 1.678 mol AsI鈧 脳 (1 mol As鈧侷鈧 / 2 mol AsI鈧) = 0.839 mol As鈧侷鈧 Now we can convert the moles of As鈧侷鈧 to mass. The molar mass of As鈧侷鈧 = 2 脳 74.92 g/mol (As) + 4 脳 126.90 g/mol (I) = 799.44 g/mol Mass of As鈧侷鈧 = 0.839 mol 脳 799.44 g/mol = 670.20 g (theoretical yield)
05

Calculate the actual mass of As鈧侷鈧 isolated

The percent yield of As鈧侷鈧 is given as 75.6%. We can use this information to find the actual mass of As鈧侷鈧 that was isolated: Actual mass of As鈧侷鈧 = (75.6 / 100) 脳 theoretical mass of As鈧侷鈧 = 0.756 脳 670.20 g = 506.51 g So, the actual mass of As鈧侷鈧 isolated is 506.51 g.

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Key Concepts

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

Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that determines the amount of product formed. It is the reactant that is completely consumed during the reaction, preventing any further product from being generated. This concept is crucial because it directly affects the theoretical yield of the reaction.

To find the limiting reactant, you first need to know the quantities of each reactant involved. By comparing the molar ratio of the reactants in the balanced chemical equation with the actual molar quantities available, you can identify which reactant will run out first. In our example, arsenic triiodide (\(\mathrm{AsI}_3\)) is the limiting reactant, as its molar ratio is less than that calculated from arsenic (\(2.061\) mol of As and \(1.678\) mol of AsI\(_3\)).
  • Determine the moles of each reactant.
  • Use the balanced equation to find the stoichiometric ratio.
  • Identify which reactant provides fewer moles of product formation (this is the limiting reactant).
Theoretical Yield
The theoretical yield is the maximum amount of product that can be formed in a reaction, assuming all of the limiting reactant is converted to product. It is calculated based on the limiting reactant using stoichiometry and the balanced chemical equation.

In our scenario, the theoretical yield of \(\mathrm{As}_{2}\mathrm{I}_{4}\) is calculated using the moles of limiting reactant, arsenic triiodide. The reaction consumes \(1.678\) mol of \(\mathrm{AsI}_3\), leading to the formation of \(0.839\) mol of \(\mathrm{As}_{2}\mathrm{I}_{4}\) given the stoichiometric equation.
  • Identify the moles of the limiting reactant.
  • Use stoichiometric calculations to determine moles of product.
  • Convert moles of product to mass using molar mass to find theoretical yield.
Percent Yield
Percent yield is a comparison of the actual yield (the amount of product actually obtained from a reaction) to the theoretical yield (the maximum possible amount of product predicted by stoichiometry). It offers insight into the efficiency of a reaction:

\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\]

In this example, your reaction had a percent yield of 75.6%. This means that 75.6% of the theoretical maximum amount of \(\mathrm{As}_{2}\mathrm{I}_{4}\) was actually collected, leading to an actual yield of \(506.51\) g.
  • Calculate theoretical yield from the limiting reactant.
  • Measure actual yield from the experiment.
  • Calculate percent yield to evaluate reaction efficiency.
Density Calculation
Density calculations are foundational in chemistry for determining the mass or volume of a substance. Density is defined as mass per unit volume and is usually expressed in \(\text{g/cm}^3\) or equivalent units.

The density formula is: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In the exercise, you need to calculate the mass of solid arsenic using its density, \(5.72\, \text{g/cm}^3\), and its volume calculated from its cubic dimensions \(3.00\, \text{cm}\) on each side, where the volume of the cube is \(27\, \text{cm}^3\). This gives a mass of arsenic of \(154.44\, \text{g}\).
  • Determine the volume of the object (especially for solids).
  • Use the density to compute the mass needed for reaction calculations.

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Most popular questions from this chapter

Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes: $$\begin{array}{c}\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\end{array}$$

The aspirin substitute. acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}\right),\) is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q)\) III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l)\) The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(C_{6} H_{5} O_{3} N\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

An element X forms both a dichloride (XCI_) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of \(10.00 \mathrm{g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{g} \mathrm{XCl}_{4} .\) Calculate the atomic mass of \(\mathrm{X},\) and identify \(\mathrm{X}\).

Boron consists of two isotopes, \(^{10} \mathbf{B}\) and \(^{11} \mathbf{B}\). Chlorine also has two isotopes, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\) How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \%\) C and \(8.16 \%\) H by mass. What is the empirical formula of this substance?

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