Question

- Butane (C4H20) reacts with oxygen (O2) according to the reaction: 2C4H10(g) + 1302(g) → 8C02(g) + 10H2O(l) L container is filled with butane to a pressure of 0.650 atm at 35°C. Oxygen is then added until the total pressure of the container reaches 1.5 atm. a. What is the partial pressure of Oxygen? b. How many moles of Butane were added to the container? C. Which reactant will limit the amount of CO2(g) formed in the reaction?

Answer 1

Given Reaction 2 CA Holal + 130 18 8co, 18) + IoHoll) Initial Pressure of container = P = 0.6south Total pressure of Container after addition of sig) - Ptotal = 1.satm PTotal = Pcatto + Pang Peale = partial pressure of Catio = 0.6soutm. Poig - Partial Pressure of 0 19) Poig = P Total - Peahio 1.satm -0.6soutm - Toug) = 0.8sum | Answer 161 Volume of containen = V = 3-SOL PV=NRI T = Temperature = 35c = (273+35)K T = 308K P = Partial Pressure of butane = 0.6scaton 0.650 atm x 3.50L = n*0.0082 L atm.moik x 308k in = o.goo7 mold Answa n = Ho of mole of Butane Partial pressure of 0.g) = 0.8 satm PV =N'RT n' = Now of moles of ₂18) | 0. 3s cm x 3.Sol = n' x 0.L cho mọi EX 302 k m' = 1.178 mole Acconding to balanced reaction 2 mole of Cahio reads with 13 mole oig) to give product

- Noo of moles of aig) required for reaction with o.goot mole cahio = (13 xo.9007) mole - 5.8smole oig) But we have only 1.178 mole og Hence, olg) will limit the amount of calgi formed / -