Gas Laws and Chemical Reactions

Question

Using stoichiometry, calculate the volume of O2 gas at 1.25 atm and 18.0 ̊C, which forms when 10.5g

KClO3 decomposes according to the reaction:

2KClO3 --> 2KCl + 3O2

Molar mass of KClO3 is 122.55 g/mol

a.2.46 L

b.0.101 L

c.0.0857 L

d.0.152 L

Answer 1

Step 1 - Write the bare (unbalanced) chemical reaction equation: KClO₃ → KCl + O₂ (we know this because Oxygen always forms as O₂ and KCl is a stable salt)

Step 2 - Balance the chemical reaction equation: 2KClO₃ → 2KCl + 3O₂

Step 3 - Using the given 10.5g of KClO₃, calculate the number of moles of Oxygen:

A) Calculate the Molar Mass of KCLO₃: Potassium = 39.098, Chlorine = 35.453, Oxygen = 3(15.999) = 47.997 and 39.098 + 35.453 + 47.997 = 122.548 g/mol.

B) Take 10.5 g and divide by 122.548 g/mol to get 0.08568 mol of KClO₃.

Step 4 - Using the 0.08568 mol of KClO₃, calculate the number of moles of O₂ gas

A) Looking at the chemical reaction equation, find the molar ration of O₂ to KClO₃. There are 3 moles of O2 for every 2 moles of KClO₃.

B) Take 0.08568 mol of KClO₃ and multiply by 3 moles of O₂ and divide by 2 moles of KClO₃ to get 0.1285 moles of O₂.

Step 5 - Convert temperature to Kelvin using K = °C + 273.15: K = 18.0 + 273.15 = 291.15K

Step 6 - Rearrange the Ideal Gas Law (PV = nRT) solving for Volume: V = nRT/P

Step 7 - Solve for V by plugging in all the numbers (and using R = 0.082057 L atm mol^-1K^-1):

V = (0.1285)(0.082057)(291.15)/(1.25) = 2.456 L = 2.46L (answer a)