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Chapter 9: Problem 122

Consider the unbalanced chemical equation \(\mathrm{KCl}+\mathrm{O}_{2} \rightarrow \mathrm{KClO}_{3}\) If you react \(42.6 \mathrm{~g}\) of \(\mathrm{KCl}\) and \(36.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and the reaction has a percent yield of \(56.0 \%\), how many grams of \(\mathrm{KClO}_{3}\) are produced?

Short Answer

Expert verified
Considering a percent yield of 56.0%, approximately 39.1 grams of potassium chlorate (KClO3) will be produced.

Step by step solution

01

Write the balanced chemical equation

First, we need to balance the given chemical equation: 2 KCl + 3 O2 -> 2 KClO3
02

Calculate the moles of reactants (KCl and O2)

Next, we will convert the given masses of KCl and O2 into moles. For this, we'll need their molar masses: Molar mass of KCl \(= 39.10~g/mol~(K) + 35.45~g/mol~(Cl) = 74.55~g/mol\) Molar mass of O2 \(= 2 × 16.00~g/mol~(O) = 32.00~g/mol\) Moles of KCl \(= \frac{42.6~g}{74.55~g/mol} = 0.571~mol\) Moles of O2 \(= \frac{36.5~g}{32.00~g/mol} = 1.14~mol\)
03

Determine the limiting reactant

To identify the limiting reactant, we will compare the mole-to-coefficient ratios of the reactants: Mole-to-coefficient ratio of KCl \(= \frac{0.571~mol}{2} = 0.285\) Mole-to-coefficient ratio of O2 \(= \frac{1.14~mol}{3} = 0.380\) Since the mole-to-coefficient ratio of KCl is smaller, KCl is the limiting reactant.
04

Calculate the theoretical yield of KClO3 in grams

We will use stoichiometry to find the theoretical yield of KClO3 based on the limiting reactant (KCl). The molar mass of KClO3 is \(= 39.10~g/mol~(K) + 35.45~g/mol~(Cl) + 3 × 16.00~g/mol~(O) = 122.55~g/mol\) For every 2 moles of KCl, 2 moles of KClO3 are produced (according to the balanced chemical equation). Therefore, we can find the moles of KClO3 produced as \(0.571~mol~(KCl) × \frac{2~mol~(KClO3)}{2~mol~(KCl)} = 0.571~mol~(KClO3)\) The theoretical yield of KClO3 in grams is: \(0.571~mol~(KClO3) × \frac{122.55~g/mol~(KClO3)}{1~mol~(KClO3)} = 69.9~g~(KClO3)\)
05

Calculate the actual yield of KClO3 considering the given percent yield

The actual yield can be calculated using the equation: Actual yield \(= \frac{Percent~Yield}{100} × Theoretical~Yield\) Actual yield of KClO3 \(= \frac{56.0\%}{100} × 69.9~g = 39.1~g\) Considering a percent yield of 56.0%, approximately 39.1 grams of potassium chlorate (KClO3) will be produced.

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Most popular questions from this chapter

Consider the unbalanced chemical equation \(\mathrm{HSbCl}_{4}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{Sb}_{2} \mathrm{~S}_{3}+\mathrm{HCl}\) If \(118.2 \mathrm{~g}\) of \(\mathrm{HSbCl}_{4}\) reacts with \(47.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and produces \(41.6 \mathrm{~g}\) of \(\mathrm{HCl}\), what is the percent yield for the reaction?

(a) Write a balanced chemical equation for the combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) Consider the reaction of three molecules of \(\mathrm{CH}_{4}(g)\) with four molecules of \(\mathrm{O}_{2}(g)\). The following boxes represent before and after pictures of the tiny, sealed container in which this reaction takes place. Draw pictures to show the contents of this container before and after the reaction happens:

Chlorine \(\left(\mathrm{Cl}_{2}\right)\) and fluorine \(\left(\mathrm{F}_{2}\right)\) react to form \(\mathrm{ClF}_{3}\). A reaction vessel contains \(10.00 \mathrm{~g} \mathrm{Cl}_{2}\) and \(10.00 \mathrm{~g} \mathrm{~F}_{2}\). (Hint: Refer to Problem 9.82.) (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(12.50 \mathrm{~g}\) of \(\mathrm{ClF}_{3}\) is recovered. What is the percent yield of the reaction?

Determine the empirical formula of the compound that is \(43.2 \%\) by mass \(\mathrm{K}, 39.1 \%\) by \(\mathrm{mass} \mathrm{Cl}\), and also contains oxygen.

A compound is \(91.77 \%\) by mass \(\mathrm{Si}\) and \(8.23 \%\) by mass \(\mathrm{H}\) and has a molar mass of approximately \(122 \mathrm{~g} / \mathrm{mol}\). What is its molecular formula?
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