A major component of gasoline is octane (C8H18). When octane is burned in air it chemically reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). What mass of oxygen gas is consumed by the reaction of 8.17g of octane?.

balance the equation:

2C8H18+25O2 >>>16CO2 + 18H2O

So for each mole of Octane you burn, you need 12.5 moles of O2

figure out the moles of octane you used, then multiply it by 12.5.
then, multiply that by 32 to get the grams of O2

To determine the mass of oxygen gas consumed by the reaction of 8.17 grams of octane (C8H18), we need to use the balanced chemical equation and the molar ratios involved in the reaction.

The balanced chemical equation for the combustion of octane is:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

From the balanced equation, we can see that in order to completely react with 2 moles of octane (C8H18), we need 25 moles of oxygen gas (O2).

Now, let's calculate the number of moles of octane in 8.17 grams using its molar mass.

Molar mass of octane (C8H18) = (12.01 g/mol x 8) + (1.01 g/mol x 18) = 114.23 g/mol

Number of moles of octane = mass / molar mass = 8.17 g / 114.23 g/mol ≈ 0.0715 mol

According to the stoichiometry of the balanced equation, 2 moles of octane require 25 moles of oxygen gas.

Therefore, if 0.0715 moles of octane require x moles of oxygen gas, we can set up the following proportion:

(0.0715 mol octane) / (2 mol octane) = (x mol oxygen) / (25 mol oxygen)

Simplifying the proportion:

0.03575 = x / 25

x = 0.03575 * 25
x ≈ 0.89375 moles of oxygen gas

Finally, to calculate the mass of oxygen gas consumed, we can use the molar mass of oxygen.

Molar mass of oxygen (O2) = 16 g/mol

Mass of oxygen gas consumed = moles of oxygen gas * molar mass of oxygen
= 0.89375 mol * 16 g/mol ≈ 14.301 g

Therefore, approximately 14.301 grams of oxygen gas is consumed by the reaction of 8.17 grams of octane.

To find the mass of oxygen gas consumed by the reaction of octane, we need to set up a balanced chemical equation for the combustion of octane.

The balanced equation for the combustion of octane is:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

According to the balanced equation, we can see that it requires 25 moles of O2 to react with 2 moles of C8H18. Therefore, the molar ratio between O2 and C8H18 is 25:2.

To find the mass of oxygen gas consumed, we need to determine the number of moles of octane first.

1. Calculate the molar mass of octane (C8H18):
Molar Mass of C (carbon) = 12.01 g/mol
Molar Mass of H (hydrogen) = 1.008 g/mol
Total Molar Mass of Octane (C8H18) = (12.01 g/mol * 8) + (1.008 g/mol * 18) = 114.23 g/mol

2. Calculate the number of moles of octane:
Number of Moles = Mass / Molar Mass
Number of Moles of Octane = 8.17 g / 114.23 g/mol ≈ 0.0715 mol

3. Determine the number of moles of O2 required using the molar ratio:
Number of Moles of O2 = Number of Moles of Octane * (25 mol O2 / 2 mol C8H18) = 0.0715 mol * (25/2) ≈ 0.894 mol

4. Calculate the mass of oxygen gas consumed:
Mass = Number of Moles * Molar Mass
Mass of Oxygen Gas Consumed = 0.894 mol * 32.00 g/mol (molar mass of O2) ≈ 28.6 g

Therefore, approximately 28.6 grams of oxygen gas is consumed by the reaction of 8.17 grams of octane.

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