Chemistry stoichiometry

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

Potentially useful information: MMKClO3=122.55 g/mol

Answer 1

2KClO₃ ==> 2KCl + 3O₂

moles KClO3 present = 10.5 g x 1 mole/122.55 g = 0.08568 moles KClO₃

From the balanced equation, 2 moles KClO₃ produce 3 moles of O₂ (stoichiometry)

Moles of O₂ produced = 0.008568 moles KClO₃ x 3 moles O₂/2 moles KClO₃ = 0.1285 moles O₂

To find the VOLUME of O₂, use the ideal gas law, PV = nRT and solve for V:

V = nRT/P = (0.1285 mol)(0.0821 L-atm/Kmol)(291 K)/1.25 atm (note to change T to Kelvin)

V = 2.46 liters (to 3 significant figures)