Question

The standard molar enthalpy of combustion of butane is -2877 kJ. C₄H₁0(g) + 13/2 O₂(g) rightarrow 4CO₂(g) + 5H₂O(g) What is the enthalpy change for the combustion of 15.00g C₄H₁0?

Answer 1

To calculate the enthalpy change for the combustion of
`15.00 g` of butane, convert the mass to moles using the molar mass of butane and multiply by the standard molar enthalpy of combustion, resulting in an enthalpy change of
`-742.7 kJ`.

Step-by-step explanation:

The student is asking how to calculate the enthalpy change for the combustion of a specific amount of butane
`(C₄H₁₀)`. Given the standard molar enthalpy of combustion of butane is
`-2877 kJ/mol`, and the balanced chemical equation is:

`C₄H₁₀(g) + 13/2 O₂(g) → 4CO₂(g) + 5H₂O(g).`

The first step to solve this problem is to determine the number of moles of butane in
`15.00 g`:

Calculate the molar mass of butane (C₄H₁₀):

`(4 × 12.01) + (10 × 1.008) = 58.14 g/mol.`

Convert grams to moles:
`15.00 g ÷ 58.14 g/mol = 0.258 moles` of butane.

Use the molar enthalpy of combustion to find the total enthalpy change:
`0.258 moles × (-2877 kJ/mol) = -742.7 kJ.`

Therefore, the enthalpy change for the combustion of
`15.00 g` of butane is
`-742.7 kJ.`