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Gasoline (which can be considered to be octane, C8H18) burns in oxygen to produce carbon dioxide and water. What volume (L) of oxygen at STP is necessary to react with 1.0 gal of gasoline? (The density of gasoline is 0.81 g/mL. 1 gal = 3.78 L)

Respuesta :

Answer:

We need 7.504 L of O2

Explanation:

Step 1: Data given

Volume of gasoline = 3.78 L

Density of gasoline = 0.81 g/mL

Molar mass octane = 114.23 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate mass of gasoline

Density = mass / volume

Mass = density * volume

Mass octane = 0.81 g/mL * 3.78 L

Mass octane = 3.0618 grams

Step 4: Calculate moles octane

Moles octane = mass octane / molar mass

Moles octane = 3.0618 grams / 114.23 g/mol

Moles octane = 0.0268 moles

Step 5: Calculate moles of O2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 0.0268 moles octane we need 12.5 * 0.0268 = 0.335 moles O2

Step 6: Calculate volume O2

1 mol = 22.4 L

0.335 moles O2 = 0.335 * 22.4 L = 7.504 L

We need 7.504 L of O2

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