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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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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 determines how much product can be formed. It is the reactant that is completely consumed first in a reaction, thus stopping the reaction from proceeding further. Understanding the concept of a limiting reactant is key in stoichiometry.

When you have a reaction with multiple reactants, it's important to determine which one will run out first. To find the limiting reactant, you must
  • Understand the stoichiometric ratios from the balanced equation.
  • Calculate how much product each reactant can potentially form.
  • Identify the reactant that yields the least product; this is the limiting reactant.
In the example exercise, when 5.0 moles of \(\mathrm{KMnO}_{4}\), \(\mathrm{KI}\), and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are mixed, \(\mathrm{KI}\) provides the least amount of product, making it the limiting reactant.
Balanced Chemical Equation
A balanced chemical equation is essential in understanding the stoichiometric relationships between reactants and products. It ensures that mass is conserved, meaning the number of atoms of each element is the same on both sides of the equation.

Consider the equation from the exercise:\[2 \mathrm{KMnO}_{4} + 10 \mathrm{KI} + 8 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4} + 2 \mathrm{MnSO}_{4} + 5 \mathrm{I}_{2} + 8 \mathrm{H}_{2} \mathrm{O}\]Features of this balanced chemical equation include:
  • The coefficients indicate the ratio in which reactants combine and products form.
  • By reading these coefficients, you can derive the stoichiometric ratios used in calculations.
  • Ensures the reaction obeys the law of conservation of mass.
Balancing an equation is a fundamental step in any stoichiometric calculation as it lays the foundation for determining the limiting reactant and potential product yields.
Mole Calculations
Mole calculations are a core component of stoichiometry. Using the balanced equation, we calculate how much product any given amount of reactant will yield. These calculations often involve converting between moles, masses, and even volumes of gases (if applicable).

For instance, in the exercise:
  • We begin by identifying the number of moles of each reactant available.
  • Use the stoichiometric coefficients from the balanced equation to compute how much product each reactant can produce.
  • Compare these quantities to determine the possible product formed and identify the limiting reactant.
  • Use the limiting reactant to calculate the actual amount of product that can form.
With these calculations, for reactants providing different quantities of product, the smallest product amount indicates both the limiting reactant and the maximum possible product yield.

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Most popular questions from this chapter

How many grams of commercial acetic acid (97\% \(\mathrm{CH}_{3} \mathrm{COOH}\) by mass) must be allowed to react with an excess of \(\mathrm{PCl}_{3}\) to produce \(75 \mathrm{g}\) of acetyl chloride \(\left(\mathrm{CH}_{3} \mathrm{COCl}\right),\) if the reaction has a \(78.2 \%\) yield? $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{PCl}_{3} & \longrightarrow \mathrm{CH}_{3} \mathrm{COCl}+ \mathrm{H}_{3} \mathrm{PO}_{3}(\text { not balanced }) \end{aligned}$$

Hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}),\) is passed over \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) at \(400^{\circ} \mathrm{C} .\) Water vapor is formed together with a black residue-a compound consisting of \(72.3 \% \mathrm{Fe}\) and \(27.7 \%\) O. Write a balanced equation for this reaction.

The manufacture of ethyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) yields diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) as a by-product. The complete combustion of a \(1.005 \mathrm{g}\) sample of the product of this process yields \(1.963 \mathrm{g} \mathrm{CO}_{2} .\) What must be the mass percents of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) and of \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) in this sample?

Nitric acid, \(\mathrm{HNO}_{3}\), can be manufactured from ammonia, \(\mathrm{NH}_{3}\), by using the three reactions shown below. $$\begin{aligned} &\text { Step 1: 4 NH }_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\\\ &\text { Step 2: } 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\\\ &\text { Step 3: } 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{aligned}$$ What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not What is the maximum number of moles of \(\mathrm{HNO}_{3}\) that can be obtained from 4.00 moles of \(\mathrm{NH}_{3}\) ? (Assume that the NO produced in step 3 is not recycled back into step 2.) (a) 1.33 mol; (b) 2.00 mol; (c) 2.67 mol; (d) 4.00 mol; (e) 6.00 mol.

How many moles of \(\mathrm{NO}(\mathrm{g})\) can be produced in the reaction of \(3.00 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g})\) and \(4.00 \mathrm{mol} \mathrm{O}_{2}(\mathrm{g}) ?\) $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

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