Question

a) The complete combustion of 10.0 g of solid naphthalene requires what volume of oxygen gas at 20.0 ^oC and 1.00 atm? How much heat energy will be transferred in this trial? Specify direction of heat transfer in your answer.

b) If the water vapor and carbon dioxide gas produced in the trial described in part d. is collected in a 20.0 L container at 427 ^oC and this container can only withstand a maximum of 5.00 atm of pressure, will the container stay intact or explode

C₁₀H₈ (s) + O₂ (g)   è CO₂ (g) + H₂O (g)    

Answer 1

Ans. #a. Balanced reaction: C₁₀H₈ + 12 O₂ ---------> 10 CO₂ + 4 H₂O

# Moles of naphthalene = Mass / MW = 10.0 g / (128.17352 g / mol) = 0.07802 mol

# Following stoichiometry of balanced reaction, the combustion of 1 mol naphthalene requires 12 mol O₂.

So, required moles of O₂ = (12 / 1) x Moles of naphthalene

                                                = 12 x 0.07802 mol = 0.93624 mol

# Now, calculate the volume of O₂ using ideal gas equation –

#b. Following stoichiometry of balanced reaction-

# Moles of CO₂ formed = 10 x moles of naphthalene = 10 x 0.07802 mol = 0.7802 mol

# Moles of H₂O formed = 4 x moles of naphthalene = 4 x 0.07802 mol = 0.31208 mol

# Total moles of gaseous products formed = 0.7802 mol + 0.31208 mol = 1.09228 mol

# Now, calculate the volume of gases formed at 5.00 atm and 427.0⁰C using ideal gas equation-

# Conclusion: Since the volume of gases formed is less than 20.0 L, the container remains intact without exploding.