In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the reaction is performed at STP?

The volume CO2 produced is 65.8 L

Explanation:

Step 1: Data given

Mass of octane = 41.9 grams

Molar mass octane = 114.23 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 41.9 grams / 114.23 g/mol

Moles octane = 0.367 moles

Step 4: Calculate moles CO2

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

For 0.367 moles octane we need 8*0.367 = 2.936 moles

Step 5: Calculate volume of CO2

1 mol = 22.4 L

2.936 moles = 22.4 * 2.936 = 65.8 L

The volume CO2 produced is 65.8 L

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