Answer to Question #149008 in General Chemistry for yem

Question

Determine the amount of theoretical oxygen and theoretical air needed to burn
a.) 60 kg of butane (C4H10)
b.) 20 kg of propane (C5H12)
c.) 80 kg of octane (C8H18)

Answer 1

A. 2C₄H₁₀ + 13O₂ ----> 8CO₂ + 10H₂O

Molar mass of C₄H₁₀ = (12x4)+10= 58gmol^-1

molar mass of O₂= 16x2=32gmol^-1

2(58)g of C₄H₁₀ required 13(32)g of Oxygen

Therefore 60Kg of C₄H₁₀ will require 416x60/116 = 215.2Kg of Oxygen

Since oxygen is 21% of Air

Theoretical amount of air needed= 100x 215.2/21 = 1025Kg of air

B. 2C₅H₁₂ + 17O₂ -----> 12CO₂ + 10H₂O

Molar mass of C₅H₁₂ = (12x5) + 12= 72gmol^-1

2(72)g of C₅H₁₂ required 17(32)g of Oxygen

Therefore, 20Kg of C₅H₁₂will require 544x20/144=75.6Kg of oxygen gas

Amount of air= 100x75.6/21= 360Kg of air

C. 2C₈H₁₈ + 25O₂ -----> 16CO₂ + 18H₂O

Molar mass of C₈H₁₈ = (12x8) + 18 = 114gmol^-1

2(114)g of C₈H₁₈ requires 25(32)g of Oxygen gas

80Kg of C₈H₁₈ will require 800x80/288 = 280.7Kg of Oxygen gas

Amount of air needed= 100x280.7/21= 1337Kg of air.