Question

19.4 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of carbon dioxide in atm is produced at 309 K in a 1.15 L flask.

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

Answer 1

Balanced equation:
2 C₄H₁₀(g) + 13 O₂(g) ===> 8 CO₂(g) + 10 H₂O(g)
Reaction type: double replacement

19.4 g of butane = 19.4 / 58.12 =  0.3337 Moles

Moles of CO2 produced =  1.335 Moles

Moles of water produced =  1.668 Moles

Total Moles = 1.335 + 1.668 = 3.003 Moles

Formula to calculate the pressure

PV = nRT

P = nRT / V

n = 3.003 Moles R = 0.0821 L atm K-1 Mol-1 T = 309 K V = 1.15 L

P = 1.335 x 0.0821 x 309 / 1.15 = 66.24 atm

Pressure of total gaes is 66.24 atm