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Chapter 4: Problem 17

A laboratory method of preparing \(\mathrm{O}_{2}(\mathrm{g})\) involves the decomposition of \(\mathrm{KClO}_{3}(\mathrm{s})\) $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) can be produced by the decomposition of \(32.8 \mathrm{g} \mathrm{KClO}_{3} ?\) (b) How many grams of \(\mathrm{KClO}_{3}\) must decompose to produce \(50.0 \mathrm{g} \mathrm{O}_{2} ?\) (c) How many grams of KCl are formed, together with \(28.3 \mathrm{g} \mathrm{O}_{2},\) in the decomposition of \(\mathrm{KClO}_{3} ?\)

Short Answer

Expert verified
The answers are: (a) 0.402 moles of \(\mathrm{O}_{2}\) can be produced from 32.8 g of \(\mathrm{KClO}_{3}\); (b) 127.4 g of \(\mathrm{KClO}_{3}\) must be decomposed to produce 50.0 g of \(\mathrm{O}_{2}\); (c) 43.9 g of \(\mathrm{KCl}\) are formed along with 28.3 g of \(\mathrm{O}_{2}\).

Step by step solution

01

Converting given grams to moles (for question a)

Calculate the amount of moles of \(\mathrm{KClO}_{3}\) from the given mass using its molar mass, which can be calculated as \(39.10 + 35.45 + (3 \times 16) = 122.55 \, \mathrm{g/mol}\).\n So, the number of moles is \(32.8 \, \mathrm{g} / 122.55 \, \mathrm{g/mol} = 0.268 \, \mathrm{mol}\).
02

Using mole-ratio to calculate moles of product

Use the stoichiometric ratio between \(\mathrm{KClO}_{3}\) and \(\mathrm{O}_{2}\) (2:3) to calculate the moles of \(\mathrm{O}_{2}\) formed. This gives \(0.268 \, \mathrm{mol} \times (3 / 2) = 0.402 \, \mathrm{mol} \, \mathrm{O}_{2}\).
03

Converting moles to grams to find required reactant mass (for question b)

Convert the given mass of \(\mathrm{O}_{2}\) into moles using its molar mass. The molar mass of \(\mathrm{O}_{2}\) is \(32.00 \, \mathrm{g/mol}\), so its moles are \(50.0 \, \mathrm{g} / 32.00 \, \mathrm{g/mol} = 1.56 \, \mathrm{mol}\). Then use the stoichiometric ratio between \(\mathrm{KClO}_{3}\) and \(\mathrm{O}_{2}\) (2:3) to find the moles of \(\mathrm{KClO}_{3}\) decomposed, which gives \(1.56 \, \mathrm{mol} \times (2 / 3) = 1.04 \, \mathrm{mol}\). The mass of \(\mathrm{KClO}_{3}\) is then calculated as \(1.04 \, \mathrm{mol} \times 122.55 \, \mathrm{g/mol} = 127.4 \, \mathrm{g}\).
04

Calculating mass of product with known mass of another product (for question c)

Convert the mass of \(\mathrm{O}_{2}\) to moles. The moles of \(\mathrm{O}_{2}\) are \(28.3 \, \mathrm{g} / 32.00 \, \mathrm{g/mol} = 0.884 \, \mathrm{mol}\). Use the stoichiometric ratio between \(\mathrm{O}_{2}\) and \(\mathrm{KCl}\) (3:2) to find the moles of \(\mathrm{KCl}\) formed, which gives \(0.884 \, \mathrm{mol} \times (2 / 3) = 0.589 \, \mathrm{mol}\). The molar mass of \(\mathrm{KCl}\) is \(74.55 \, \mathrm{g/mol}\), so the mass of \(\mathrm{KCl}\) is \(0.589 \, \mathrm{mol} \times 74.55 \, \mathrm{g/mol} = 43.9 \, \mathrm{g}\).

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Most popular questions from this chapter

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

An organic liquid is either methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) or a mixture of the two. A 0.220-g sample of the liquid is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) and yields \(0.352 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) .\) Is the liquid a pure alcohol or a mixture of the two?

Briefly describe (a) balancing a chemical equation; (b) preparing a solution by dilution; (c) determining the limiting reactant in a reaction.

A seawater sample has a density of \(1.03 \mathrm{g} / \mathrm{mL}\) and \(2.8 \% \mathrm{NaCl}\)l by mass. A saturated solution of \(\mathrm{NaCl}\) in water is \(5.45 \mathrm{M} \mathrm{NaCl} .\) How many liters of water would have to be evaporated from \(1.00 \times 10^{6} \mathrm{L}\) of the seawater before \(\mathrm{NaCl}\) would begin to crystallize? (A saturated solution contains the maximum amount of dissolved solute possible.)
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