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Chapter 4: Problem 137

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

Short Answer

Expert verified
The maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}\), \(\mathrm{KI}\), and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is 3.0 moles. Hence, the correct option is (a) 3.0 \(\mathrm{mol}\).

Step by step solution

01

Identify the Stoichiometric Ratios

From the balanced chemical equation, for every 2 moles of \(\mathrm{KMnO}_{4}\), 10 moles of \(\mathrm{KI}\), and 8 moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) react to form 6 moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\). The stoichiometric ratios between these reactants and the product are thus 2:10:8:6 for \(\mathrm{KMnO}_{4}\), \(\mathrm{KI}\), \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) respectively.
02

Determine the Limiting Reactant

In the given problem, the mixture contains 5.0 moles each of \(\mathrm{KMnO}_{4}\), \(\mathrm{KI}\), and \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be produced from each reactant based on their stoichiometric ratios are 5.0 * (6/2) = 15 moles from \(\mathrm{KMnO}_{4}\), 5.0 * (6/10) = 3.0 moles from \(\mathrm{KI}\), and 5.0 * (6/8) = 3.75 moles from \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The limiting reactant is the one that produces the least amount of the product, which in this case is \(\mathrm{KI}\).
03

Calculate the Maximum Moles of Product

As \(\mathrm{KI}\) is the limiting reactant, the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be produced is the amount \(\mathrm{KI}\) can produce. So, the maximum moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is 3.0.

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Most popular questions from this chapter

Iron ore is impure \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) When \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is heated with an excess of carbon (coke), metallic iron and carbon monoxide gas are produced. From a sample of ore weighing \(938 \mathrm{kg}, 523 \mathrm{kg}\) of pure iron is obtained. What is the mass percent \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) by mass, in the ore sample, assuming that none of the impurities contain Fe?

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

In the reaction of \(2.00 \mathrm{mol} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\), \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. $$ \mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl} $$ (a) The theoretical yield is \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (b) The theoretical yield is \(1.00 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (c) The theoretical yield depends on how large an excess of HF is used. (d) The percent yield is \(85 \%\)

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?
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