When it is heated, potassium chlorate decomposes to potassium chloride and oxygen: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) A sample of solid \(\mathrm{KClO}_{3}\) decomposes to produce \(150.0 \mathrm{~mL}\) of oxygen at \(22.0^{\circ} \mathrm{C}\) and \(780.5 \mathrm{~mm} \mathrm{Hg}\). How many grams of \(\mathrm{KClO}_{3}\) decomposed?

Short Answer

Expert verified
0.511 grams of potassium chlorate decomposed to produce 150.0 mL of oxygen gas at the given conditions.

Step by step solution

01

Convert the gas volume to moles

To solve this problem, we first need to convert the given volume of oxygen gas to moles. We will use the ideal gas law formula, which is: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We are given the pressure P = 780.5 mmHg, the volume V = 150.0 mL, and the temperature T = 22.0°C. First, let's convert the given temperature and pressure to the appropriate units: Finally, convert the volume from milliliters to liters: \(V = 150.0 \ mL = 0.150 \ L\) Temperature: \(T = 22.0^{\circ}C + 273.15 = 295.15\ K\) Pressure: \(P = 780.5 \ mmHg \times \frac{1 \ atm}{760 \ mmHg} = 1.027 \ atm\) Now, we can rearrange the ideal gas law formula to solve for n (moles of O₂): \(n = \frac{PV}{RT}\)
02

Calculate moles of oxygen gas produced

Plug the given values into the equation, we get: \(n_{O_{2}} = \frac{1.027 \times 0.150}{0.0821 \times 295.15} \) \(n_{O_{2}} = 0.00626 \ mol\) So, 0.00626 moles of oxygen gas were produced during the decomposition.
03

Use stoichiometry to find moles of potassium chlorate

From the balanced chemical equation, we can see that 2 moles of potassium chlorate decompose to produce 3 moles of oxygen gas. We can set up a proportion to find the number of moles of potassium chlorate: \(\frac{2 \ moles \ KClO_{3}}{3 \ moles \ O_{2}} = \frac{moles \ KClO_{3}}{0.00626 \ moles \ O_{2}}\) Now, solve for moles of KClO₃: \(moles \ KClO_{3} = 2 \times \frac{0.00626}{3}\) \(moles \ KClO_{3} = 0.00417 \ mol\)
04

Calculate the mass of potassium chlorate

Finally, we can find the mass of potassium chlorate by multiplying the number of moles by the molar mass of the compound. The molar mass of potassium chlorate is: \(KClO_{3} = 39.10 (K) + 35.45 (Cl) + 3(16.00(O)) = 122.55 \ g/mol\) Multiply the moles of KClO₃ by its molar mass: \(mass \ KClO_{3} = 0.00417 \ mol \times 122.55 \ \frac{g}{mol}\) \(mass \ KClO_{3} = 0.511 \ g\) So, 0.511 grams of potassium chlorate decomposed to produce 150.0 mL of oxygen gas at the given conditions.

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