If 5.45 g of potassium chlorate decompose, how many liters of oxygen gas are given off at 1.58 atm and 32 degrees C? 2KClO3 → 2KCl + 3O2

1.04L

Explanation:

First, we'll begin by calculating the number of mole of KClO3 present in 5.45g of the compound.

This is illustrated below:

Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 39 + 35.5 + 48 = 122.5g/mol

Mass of KClO3 from the question = 5.45g

Mole of KClO3 = ?

Number of mole = Mass / Molar Mass

Mole of KClO3 = 5.45/122.5

Mole of KClO3 = 0.044mole

The equation for the reaction is given below:

2KClO3 → 2KCl + 3O2

From the equation above,

2 moles of KClO3 produced 3 moles of O2.

Therefore, 0.044 mole of KClO3 will produce = (0.044 x 3) / 2 = 0.066 mole of O2.

Now, we can obtain the volume of O2 given off by using the ideal gas equation as illustrated below:

Data obtained from the question include:

P (pressure) = 1.58 atm

T (temperature) = 32°C = 32 + 273 = 305K

V (volume) = ?

n (number of mole of O2) = 0.066 mole

R (gas constant) = 0.082atm. L/Kmol

Using the ideal gas equation PV = nRT, the volume of O2 given off can be obtained as follow:

PV = nRT

1.58 x V = 0.066 x 0.082 x 305

Divide both side by 1.58

V = (0.066 x 0.082 x 305) / 1.58

V = 1.04L

Therefore, the volume of O2 given off is 1.04L