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Chapter 6: Problem 51

A 3.57 g sample of a \(\mathrm{KCl}-\mathrm{KClO}_{3}\) mixture is decomposed by heating and produces \(119 \mathrm{mL} \mathrm{O}_{2}(\mathrm{g})\) measured at \(22.4^{\circ} \mathrm{C}\) and \(738 \mathrm{mm} \mathrm{Hg}\). What is the mass percent of \(\mathrm{KClO}_{3}\) in the mixture? $$2 \mathrm{KClO}_{3}(\mathrm{s}) \longrightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g})$$

Short Answer

Expert verified
The mass percent of \(KClO_3\) in the mixture is 11.1%.

Step by step solution

01

Convert given conditions to standard form

First, it's important to remember that the volume should be in liters and the pressure should be in atm for the ideal gas laws equation. The given volume is 119 mL, which can be converted to liters by dividing by 1000 (119 mL = 0.119 L), and the given pressure is 738 mm Hg, which can be converted to atmospheres by dividing by 760 (738 mmHg = 0.971 atm). The temperature is given in degree celsius and it needs to be converted to Kelvins (K = degree celsius + 273). Thus, the temperature is 22.4 + 273 = 295.4 K.
02

Use ideal gas law to find moles of O2

Use the ideal gas equation \(PV = nRT\), where P is the pressure (0.971 atm), V is the volume (0.119 L), T is the temperature (295.4 K), n is the number of moles (which we are trying to find), and R is the ideal gas constant (0.0821 L·atm/mol·K). Solve the equation for n: \(n = \frac{PV}{RT} = \frac{(0.971 atm)(0.119 L)}{(0.0821 L⋅atm/mol⋅K)(295.4 K)} = 0.00484 mol\) of \(O_2\).
03

Use stoichiometry to get moles of KClO3

We know that from the balanced chemical equation provided, 2 moles of \(KClO_3\) yields 3 moles of \(O_2\). Thus, converting the moles of \(O_2\) to moles of \(KClO3\), we get \(n_{KClO3} = 0.00484 mol O2 * 2/3 = 0.00323 mol of KClO3\).
04

Calculate mass of KClO3

Multiply the obtained moles of \(KClO_3\) by the molar mass of \(KClO_3\) (122.55 g/mol) to get the mass: \(mass_{KClO3} = 0.00323 mol * 122.55 g/mol = 0.396 g\)
05

Calculate mass percent of KClO3

Finally, apply the mass percent formula, \(mass percent = \frac{mass of part}{mass of whole} \times 100%\). In this context, it will be \(mass percent_{KClO3} = \frac{0.396 g}{3.57 g} \times 100% = 11.1%\)

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Most popular questions from this chapter

A particular application calls for \(\mathrm{N}_{2}(\mathrm{g})\) with a density of \(1.80 \mathrm{g} / \mathrm{L}\) at \(32^{\circ} \mathrm{C} .\) What must be the pressure of the \(\mathrm{N}_{2}(\mathrm{g})\) in millimeters of mercury? What is the molar volume under these conditions?

What is the volume, in liters, occupied by a mixture of 15.2 \(\mathrm{g} \mathrm{Ne}(\mathrm{g})\) and \(34.8 \mathrm{g} \mathrm{Ar}(\mathrm{g})\) at 7.15 atm pressure and \(26.7^{\circ} \mathrm{C} ?\)

A 0.418 g sample of gas has a volume of \(115 \mathrm{mL}\) at \(66.3^{\circ} \mathrm{C}\) and \(743 \mathrm{mm} \mathrm{Hg} .\) What is the molar mass of this gas?

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is used to disinfect contact lenses. How many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) can be liberated from \(10.0 \mathrm{mL}\) of an aqueous solution containing \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass? The density of the aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.01 \mathrm{g} / \mathrm{mL}\) $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{O}_{2}(\mathrm{g})$$
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