Chemistry Problem Solving: Decomposition of Potassium Chlorate

How many grams of solid potassium chlorate (KClO3) decompose to form solid potassium chloride and 790 mL of oxygen gas at 111 °C and 770 torr? (3 sf)

Answer: 1.95 g KClO3

Answer:

Given the conditions provided, we can determine the mass of potassium chlorate (KClO3) that decomposes to form potassium chloride and oxygen gas.

First, we calculate the moles of oxygen gas (O2) formed using the ideal gas equation:

P = 770 torr = 0.983 atm

V = 790 mL = 0.790 L

T = 111 °C + 273 = 384 K

Using PV = nRT, we find n = 0.0238 mol O2

Next, we use the coefficients in the chemical equation to determine the moles of KClO3 reacted: 0.0159 mol KClO3

Finally, by calculating the molar mass of KClO3 (122.55 g/mol), we obtain the mass of KClO3 that decomposes to be 1.95 g.

In this chemistry problem, we are tasked with finding the amount of potassium chlorate (KClO3) that decomposes to produce potassium chloride and oxygen gas under specific conditions.

By applying the ideal gas equation and utilizing the coefficients of the chemical equation, we were able to calculate the mass of KClO3 that reacts to form the given volume of oxygen gas.

The process involves converting pressure to atm, volume to L, and temperature to Kelvin to facilitate the calculations accurately. Furthermore, understanding the stoichiometry of the reaction is crucial in determining the correct amount of reactant consumed.

Overall, this problem showcases the application of gas laws and stoichiometry in determining the quantity of reactants and products in a chemical reaction.