Butane (C4H10) burns in the presence of oxygen to produce carbon dioxide gas and water vapor. Write a balanced equation for this reaction and calculate the number of liters of carbon dioxide measured at STP that could be produced from 5.69g of butane

8.78 L

Explanation:

The balanced equation for the combustion of butane (C₄H₁₀) in presence of oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is the following:

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O

First, we calculate the moles there are in 5.69 g of butane, by dividing the mass into the molecular weight (MM butane= 58 g/mol):

moles C₄H₁₀= mass/MM= 5.69 g/(58 g/mol)= 0.098 mol

According to the equation, 2 moles of C₄H₁₀ produce 8 moles of CO₂. Thus, if we have 0.098 moles C₄H₁₀:

moles CO₂= 0.098 mol C₄H₁₀ x 8 mol CO₂/(2 mol C₄H₁₀)= 0.392 mol CO₂

Finally, we know that 1 mole of any gas occupies a volume of 22.4 L measured at STP, so we calculate the volume of CO₂ as follows:

volume CO₂= 0.392 mol CO₂ x 22.4 L/1 mol CO₂= 8.78 L

it should be p1 v1 = p2 v2

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explanation:

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