Given the equation 2C8H18 + 25O2 = 16 CO2 + 18H2O: How many moles of oxygen gas (O2) are consumed upon combustion of 2.20 L of octane ( D = 0.703 g/mL)?
yes
To determine the number of moles of oxygen gas (O2) consumed upon combustion of octane (C8H18), we need to convert the volume of octane to moles.
First, we need to calculate the mass of octane using its density and volume.
Density (D) = 0.703 g/mL
Volume = 2.20 L = 2200 mL
Mass of octane (C8H18) = Density x Volume
= 0.703 g/mL x 2200 mL
= 1544.6 g
Next, we need to convert the mass of octane to moles. We can use the molar mass of octane to do this.
Molar mass of octane (C8H18):
C: 8 atoms x atomic mass (12.01 g/mol) = 96.08 g/mol
H: 18 atoms x atomic mass (1.01 g/mol) = 18.18 g/mol
Total molar mass of octane = 96.08 g/mol + 18.18 g/mol
= 114.26 g/mol
Now we can calculate the number of moles of octane:
Moles = Mass / Molar mass
= 1544.6 g / 114.26 g/mol
= 13.51 mol
Based on the balanced chemical equation, each molecule of octane consumes 25 molecules of oxygen gas (O2).
Therefore, the number of moles of O2 consumed = 13.51 mol x 25 mol/mol
= 337.75 mol
So, 337.75 moles of oxygen gas (O2) are consumed upon the combustion of 2.20 L of octane.
To find the number of moles of oxygen gas consumed, we need to use the balanced chemical equation and the given amount of octane.
First, let's calculate the mass of octane (C8H18):
Mass = volume × density
Mass = 2.20 L × 0.703 g/mL = 1.546 g
Next, we convert the mass of octane to moles using its molar mass:
Molar mass of octane (C8H18) = 12.01 g/mol × 8 + 1.01 g/mol × 18 = 114.23 g/mol
Moles of octane = 1.546 g / 114.23 g/mol = 0.0135 mol
Now, we can use the balanced chemical equation to determine the mole ratio between octane and oxygen gas:
2C8H18 + 25O2 = 16CO2 + 18H2O
From the equation, we can see that 25 moles of O2 react with 2 moles of C8H18.
Therefore, the moles of O2 consumed = (25 mol O2 / 2 mol C8H18) × 0.0135 mol C8H18
= 0.16875 mol O2
So, approximately, 0.16875 moles of oxygen gas (O2) are consumed upon combustion of 2.20 L of octane.
well, since all the compounds are gases, you can use the law of volumes to surmise
2*1.1 liters octane
25*1.1 liters oxygen
16*1.1 liters carbon dioxide
18*1.1 liters water vapor.
So moles of O2? it will be (25*1.1)/22.4 moles O2. No need to worry about temp, pressure, because all the gases are at same temp, pressure.