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Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 4.3 g of hexane is mixed with 25.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

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Answer:

13g of CO₂ is the maximum amount that could be produced

Explanation:

The reaction of hexane with oxygen is:  

C₆H₁₄ + 19/2O₂ → 6CO₂ + 7H₂O  

Where 19/2 moles of oxygen react per mole of C₆H₁₄

 

To solve this question we need to find theoretical yield finding limiting reactant :

Moles C₆H₁₄:  

4.3g C₆H₁₄ * (1mol / 86.18g) = 0.0499 moles  

Moles O₂:  

25.8g * (1mol / 32g) = 0.806 moles  

For a complete reaction of 0.0499 moles of C₆H₁₄ are needed:

0.0499 moles of C₆H₁₄ * (19/2 mol O₂ / 1mol C₆H₁₄) = 0.474 moles of O₂.

As there are 0.806 moles, O₂ is in excess and C₆H₁₄ is limiting reactant

In theoretical yield, the moles of hexane added = 6Moles of CO₂ produced. Moles of CO₂ are:

0.0499 moles C₆H₁₄ * (6mol CO₂ / 1mol C₆H₁₄) = 0.299 moles CO₂

In grams:

0.299 moles CO₂ * (44.01g / mol) = 13g CO₂

13g of CO₂ is the maximum amount that could be produced

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