Chemistry Practice Problem Help?? (University)

Question

Q: 1.00 L of octane (C8H18, M=114.23 g/mol, d= 0.703 g/cm^3) undergoes a combustion reaction with 5.00L of oxygen at 25 degrees celcius and 1.00 atm pressure. What mass of carbon dioxide (M=44.01 g/mol) is produced?

R=0.083145 Lbar mol^-1 K^-1

1 atm= 1.01325 bar

Answer 1

2 C₈H₁₈ + 25 O₂ ==> 16 CO₂ + 18 H₂O ... balanced equation for combustion of octane

Let us first find if octane or oxygen is in limiting supply:

moles octane present = 1.00 L x 703 g/L x 1 mol/114.23 g = 6.15 moles C₈H₁₆ present

moles oxygen present: PV=nRT and n=PV/RT.

P = 1.00 atm x 1.01325 bar/atm = 1.01325 bar

V = 5.00 L

R = 0.083145 Lbar/Kmol

T = 25 + 273 = 298K

n = (1.01325)(5.00)/(0.083145)(298) = 0.204 moles O₂ present

Clearly, O₂ is limiting.

Next, from the moles of O2 present, we can find moles of CO₂ produced:

0.204 moles O₂ x 16 moles CO₂/25 moles O₂ = 0.1306 moles CO₂ produced

Next, convert moles of CO₂ to mass of CO₂:

0.1306 moles CO₂ x 44.01 g/mole = 5.75 g CO₂ (to 3 significant figures)