A mixture containing $\mathrm{KClO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\( and \)\mathrm{KCl}\( was heated, producing \)\mathrm{CO}_{2}, \mathrm{O}_{2}\(, and \)\mathrm{H}_{2} \mathrm{O}$ gases according to the following equations: $$ \begin{aligned} 2 \mathrm{KClO}_{3}(s) & \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \\\ 2 \mathrm{KHCO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\ \mathrm{K}_{2} \mathrm{CO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$ The KCl does not react under the conditions of the reaction. If $100.0 \mathrm{~g}\( of the mixture produces \)1.80 \mathrm{~g}\( of \)\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{~g}\( of \)\mathrm{CO}_{2}$, and \(4.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Step by step solution

01

Identify the corresponding coefficients in the balanced equations

In the balanced chemical equations, the coefficients represent the stoichiometric ratio of each reactant and product. We will use these coefficients to relate the amount of reactants in the original mixture to the amount of products formed. For \(\mathrm{KClO}_{3}(s)\): 2 moles of \(\mathrm{KClO}_{3}\) produce 3 moles of \(\mathrm{O}_{2}\): $$2 \mathrm{KClO}_{3} \longrightarrow 3 \mathrm{O}_{2}$$ For \(\mathrm{KHCO}_{3}(s)\): 2 moles of \(\mathrm{KHCO}_{3}\) produce 1 mole of \(\mathrm{H}_{2} \mathrm{O}\) and 2 moles of \(\mathrm{CO}_{2}\): $$2 \mathrm{KHCO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O} + 2 \mathrm{CO}_{2}$$ For \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\): 1 mole of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) produces 1 mole of \(\mathrm{CO}_{2}\): $$\mathrm{K}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{CO}_{2}$$

02

Obtain the moles of each gas produced in the given mixture

First, we need to find the moles of \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) produced from the given mass of each gas: - Moles of \(\mathrm{CO}_{2}\): $$\dfrac{13.20~g}{44.01\frac{g}{mol}} = 0.3~mol\,\, \mathrm{CO}_{2}$$ - Moles of \(\mathrm{O}_{2}\): $$\dfrac{4.00~g}{32.00\frac{g}{mol}} = 0.125~mol\,\, \mathrm{O}_{2}$$ - Moles of \(\mathrm{H}_2 \mathrm{O}\): $$\dfrac{1.80~g}{18.02\frac{g}{mol}} = 0.1~mol\,\, \mathrm{H}_2 \mathrm{O}$$

03

Determine the amount of reacting compounds in the original mixture

Use the stoichiometric ratios of the balanced chemical equations to determine the moles of \(\mathrm{KClO}_{3}, \mathrm{KHCO}_{3}\), and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) in the original \(100.0 \mathrm{~g}\) mixture: - From \(\mathrm{KClO}_{3}\): $$\dfrac{0.125~mol\,\, \mathrm{O}_{2}}{3/2} = 0.0833~mol\, \mathrm{KClO}_{3}$$ - From \(\mathrm{KHCO}_{3}\): $$\dfrac{0.1~mol \,\mathrm{H}_{2} \mathrm{O}}{1/2} = 0.2~mol\, \mathrm{KHCO}_{3}$$ - From \(\mathrm{K}_{2} \mathrm{CO}_{3}\): $$\dfrac{0.3~mol\,\, \mathrm{CO}_{2}}{2 + 1} = 0.1~mol\, \mathrm{K}_{2} \mathrm{CO}_{3}$$

04

Calculate the mass of each compound in the original mixture

Now, determine the mass of each compound in the original mixture using their molar masses: - Mass of \(\mathrm{KClO}_{3}\): $$0.0833~mol \times 122.55\frac{g}{mol} = 10.21~g\, \mathrm{KClO}_{3}$$ - Mass of \(\mathrm{KHCO}_{3}\): $$0.2~mol \times 100.12\frac{g}{mol} = 20.02~g\, \mathrm{KHCO}_{3}$$ - Mass of \(\mathrm{K}_{2} \mathrm{CO}_{3}\): $$0.1~mol \times 138.21\frac{g}{mol} = 13.82~g\, \mathrm{K}_{2} \mathrm{CO}_{3}$$

05

Calculate the mass of \(\mathrm{KCl}\) in the original mixture

Subtract the total mass of the other compounds from \(100.0 \mathrm{~g}\) to find the quantity of \(\mathrm{KCl}\): Mass of \(\mathrm{KCl}\): $$100.0~g - 10.21~g - 20.02~g - 13.82~g = 55.95~g\, \mathrm{KCl}$$

06

Write the final composition of the original mixture

The composition of the original \(100.0 \mathrm{~g}\) mixture is: - \(10.21~g\, \mathrm{KClO}_{3}\) - \(20.02~g\, \mathrm{KHCO}_{3}\) - \(13.82~g\, \mathrm{K}_{2} \mathrm{CO}_{3}\) - \(55.95~g\, \mathrm{KCl}\)

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