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Chapter 1: Problem 12

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g

Short Answer

Expert verified
The amount of oxygen gas produced when 12.25 g of potassium chlorate decomposes is (C) 4.80 g.

Step by step solution

01

Convert the mass of potassium chlorate to moles

To start with the stoichiometric calculations first need to convert the given mass of potassium chlorate (\(\mathrm{KClO}_3\)) into moles. Moles can be calculated by the mass of the substance divided by its molar mass. From the periodic table, the molar mass of potassium chlorate \(\mathrm{KClO}_3\) is 39.10 (potassium) + 35.45 (chlorine) + 3*16.00 (oxygen) = 122.55 g/mol. So, the moles of \(\mathrm{KClO}_3\) used is 12.25 g / 122.55 g/mol = 0.100 mol.
02

Use stoichiometry to find moles of oxygen

Now, using the balanced chemical equation, we find the stoichiometric ratio between potassium chlorate and oxygen. The equation is \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\). This tells us that for every 2 moles of \(\mathrm{KClO}_3\) that react, 3 moles of oxygen gas \(\mathrm{O}_2\) are produced. So the moles of \(\mathrm{O}_2\) produced are \(0.100\, \mathrm{mol} \times \frac{3\, \mathrm{mol\, O_2}}{2\, \mathrm{mol\, KClO}_3} = 0.150\, \mathrm{mol}\).
03

Convert the moles of oxygen to grams

Finally, convert the moles of oxygen to grams to find out how much oxygen will be produced. The molar mass of oxygen is (16.00*2) = 32.00 g/mol. So, the mass of \(\mathrm{O}_2\) produced is \(0.150\, \mathrm{mol} \times 32.00\, \mathrm{g/mol} = 4.80\, \mathrm{g}\). So, the correct answer is (C) 4.80 g.

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Most popular questions from this chapter

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

A 1 -molar solution of a very weak monoprotic acid has a pH of 5. What is the value of \(K_{a}\) for the acid? (A) \(K_{a}=1 \times 10^{-10}\) (B) \(K_{a}=1 \times 10^{-7}\) (C) \(K_{a}=1 \times 10^{-5}\) (D) \(K_{a}=1 \times 10^{-2}\)

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)
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