Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

How many moles of butane (C4H10(g)) were combusted in the presence of excess
oxygen gas if 55.8 L of water vapour at STP is collected?

Sagot :

ardni313

Moles of Butane combusted : 0.4982

Further explanation

Given

55.8 water vapour at STP

Required

moles of Butane

Solution

Reaction(Combustion of Butane) :

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

1 mol at STP = 22.4 L, so mol of H₂O for 55.8 L :

=55.8 : 22.4

=2.491 moles

mol H₂O based on Butane as a limiting reactant(excess  Oxygen)

From the equation, mol ratio Butane-C₄H₁₀ : H₂O = 2 : 10, so mol Butane :

[tex]\tt =\dfrac{2}{10}\times 2.491\\\\=0.4982[/tex]

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.