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

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 :

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.

listen, you cant how young you were.

"you didn't chose the stud life. the stud life chose you."