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?

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 :

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic