Question

When potassium chlorate decomposes, how many g of KCLO3 are needed to produce 659 L of oxygen at 785 mmHg and 30.0 C?

Answer 1

Answer

2229.32 grams

Explanation

Given:

Volume of oxygen produced = 659 L

Pressure, P = 785 mmHg = 785/760 = 1.03 atm

Temperature, T = 30.0 °C = (30.0 +273.15 K) = 303.15 K

What to find:

The mass in grams of KClO3 that decompose.

Step-by-step solution:

The question asked to find the mass, in g, of KClO3 that decomposes to give a certain 659 L of O2.

Step 1: Write a balanced chemical equation for the reaction:

2KClO3 (s) ----> 2KCl (s) + 3O2 (g)

Step 2: Use the ideal gas equation to find the moles of O2 that form.

The ideal gas equation is:

$\begin{gathered} PV=nRT \\ \\ \Rightarrow n=\frac{PV}{RT} \end{gathered}$

Plugging in known values, and R = 0.082057 L.atm/mol.K we have:

$n=\frac{1.03\text{ }atm\times659\text{ }L}{(0.082057\text{ }L.atm\text{/}mol.K)\times303.15\text{ }K}=27.2866\text{ }mol$

Step 3: Use the coefficients of the chemical equation to find the relative number of moles of KClO3 that reacted:

From the balanced equation, 2 moles of KClO3 produced 3 moles of O2

Therefore, the moles of KClO3 that will produce 27.2866 moles of O2 is:

$\frac{2\text{ }mol\text{ }KClO_3\times27.2866\text{ }mol\text{ }O_2}{3\text{ }mol\text{ }O_2}=18.1911\text{ }mol\text{ }KClO_3$

Finally, use the molar mass of potassium chlorate (122.55 g/mol) to find the number of grams that reacted:

$\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Mass=Moles\times Molar\text{ }mass \\ \\ Mass=18.1911\text{ }mol\times122.55\text{ }g\text{/}mol \\ \\ Mass=2229.32\text{ }grams \end{gathered}$

2229.32 grams of KClO3 are needed to produce 659 L of oxygen at 785 mmHg and 30.0 °C.