/*! 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 56 The copper(II) ion in a copper(I... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 56

The copper(II) ion in a copper(II) sulfate solution reacts with potassium iodide to produce the triiodide ion, \(\mathrm{I}_{3}^{-}\). This reaction is commonly used to determine how much copper is present in a given sample. $$\operatorname{CuSO}_{4}(a q)+\operatorname{KI}(a q) \rightarrow \operatorname{CuI}(s)+\mathrm{KI}_{3}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)$$ If \(2.00 \mathrm{g}\) of \(\mathrm{KI}\) is added to a solution containing 0.525 \(\mathrm{g}\) of \(\mathrm{CuSO}_{4},\) calculate the mass of each product produced.

Short Answer

Expert verified
The mass of each product produced is as follows: \(CuI = 0.626 \, g\), \(KI_3 = 1.29 \, g\), and \(K_2 SO_4 = 0.573 \, g\).

Step by step solution

01

Write the balanced chemical equation

First, write the balanced chemical equation for the reaction: \(CuSO_4(aq) + 2KI(aq) \rightarrow CuI(s) + KI_3(aq) + K_2SO_4(aq)\) This equation tells us the stoichiometric ratios between the reactants and products: 1 mole of \(CuSO_4\) reacts with 2 moles of \(KI\) to produce 1 mole of \(CuI\), 1 mole of \(KI_3\), and 1 mole of \(K_2SO_4\).
02

Determine the limiting reactant

Now, we need to determine which reactant is limiting. Convert the given masses of reactants to moles using their molar masses: For \(CuSO_4\): mass = 0.525 g, molar mass = 159.61 g/mol Moles of \(CuSO_4\) = (0.525 g) / (159.61 g/mol) = 0.00329 mol For \(KI\): mass = 2.00 g, molar mass = 166.0 g/mol Moles of \(KI\) = (2.00 g) / (166.0 g/mol) = 0.0120 mol Now, divide the moles of each reactant by their stoichiometric coefficients, and find the smallest value: \(CuSO_4\): 0.00329 mol / 1 = 0.00329 \(KI\): 0.0120 mol / 2 = 0.00600 Since 0.00329 is smaller than 0.006, \(CuSO_4\) is the limiting reactant.
03

Calculate the mass of each product

Based on the stoichiometric ratios and the limiting reactant (\(CuSO_4\)), we can calculate the mass of each product produced: 1. Mass of \(CuI\) produced: Moles of \(CuI\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(CuI\) = moles 脳 molar mass = (0.00329 mol) 脳 (190.45 g/mol) = 0.626 g 2. Mass of \(KI_3\) produced: Moles of \(KI_3\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(KI_3\) = moles 脳 molar mass = (0.00329 mol) 脳 (391.82 g/mol) = 1.29 g 3. Mass of \(K_2SO_4\) produced: Moles of \(K_2SO_4\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(K_2SO_4\) = moles 脳 molar mass = (0.00329 mol) 脳 (174.26 g/mol) = 0.573 g So, the mass of \(CuI\) produced is 0.626 g, the mass of \(KI_3\) produced is 1.29 g, and the mass of \(K_2SO_4\) produced is 0.573 g.

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Key Concepts

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

Chemical Reaction
A chemical reaction involves the transformation of one set of chemical substances into another. It happens through the breaking and forming of chemical bonds, leading to changes in the composition of matter. The reaction between copper(II) sulfate and potassium iodide to produce triiodide ion is one such chemical transformation.

It's crucial for students to recognize that chemical equations provide a concise way of demonstrating the conversion of reactants to products. The balanced chemical equation for this reaction is fundamental to understanding the stoichiometry, as it dictates the exact proportions in which the substances react and form products.
Limiting Reactant
The concept of the limiting reactant is essential in stoichiometry. It determines the amount of product that can be formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first, which causes the reaction to stop. Once the limiting reactant is depleted, no more product can form, regardless of how much of the other reactants remain.

In the exercise provided, understanding which reactant is limiting involves comparing the mole ratio of the reactants with the coefficients in the balanced equation. By calculating the molar amounts and dividing by the stoichiometric coefficients, students can easily identify the limiting reactant, which in this case is copper(II) sulfate ((CuSO_4)).
Molar Mass
Molar mass, a vital concept in chemistry, is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It directly relates the mass of a substance to the number of moles present, making it indispensable in converting between mass and moles for chemical calculations.

When solving stoichiometry problems, like the copper(II) sulfate-potassium iodide reaction, students must use the molar mass to convert the given mass of each reactant to moles. It's important to extract the correct molar masses from the periodic table for each chemical compound involved in the reaction, as any error in these values can lead to incorrect outcomes in determining the limiting reactant and the mass of products formed.

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Most popular questions from this chapter

Magnesium metal, which burns in oxygen with an intensely bright white flame, has been used in photographic flash units. The balanced equation for this reaction is $$2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)$$ How many grams of \(\mathrm{MgO}(s)\) are produced by complete reaction of \(1.25 \mathrm{g}\) of magnesium metal?

For each of the following unbalanced equations, calculate the mass of each product that could be produced by complete reaction of \(1.55 \mathrm{g}\) of the reactant indicated in boldface. a. \(\mathbf{C S}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathbf{N a N O}_{3}(s) \rightarrow \operatorname{NaNO}_{2}(s)+\mathrm{O}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathbf{M} \mathbf{n} \mathbf{O}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathbf{B} \mathbf{r}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{BrCl}(g)\)

For each of the following unbalanced equations, calculate how many moles of the second reactant would be required to react completely with exactly \(25.0 \mathrm{g}\) of the first reactant. Indicate clearly the mole ratio used for each conversion. a. \(\mathrm{Mg}(s)+\mathrm{CuCl}_{2}(a q) \rightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{Cu}(s)\) b. \(\operatorname{AgNO}_{3}(a q)+\mathrm{NiCl}_{2}(a q) \rightarrow \operatorname{AgCl}(s)+\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) c. \(\mathrm{NaHSO}_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{KHCO}_{3}(a q)+\mathrm{HCl}(a q) \rightarrow\) \(\mathrm{KCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\)

Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst. $$\mathrm{Xe}(g)+2 \mathrm{F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s)$$ What is the theoretical mass of xenon tetrafluoride that should form when \(130 . \mathrm{g}\) of xenon is reacted with \(100 . \mathrm{g}\) of \(\mathrm{F}_{2}\) ? What is the percent yield if only \(145 \mathrm{g}\) of \(\mathrm{XeF}_{4}\) is actually isolated?

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Calculate the mass of each product gas if \(1.25 \mathrm{g}\) of ammonium nitrate reacts.
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