Question

When 2.50 g of methane burns in oxygen, 125 kj of heat is produced. what is the enthalpy of combustion per mole of methane under these conditions?

Answer 1

MCH₄: 16 g/mol
..............................
CH₄ + 2O₂ ---> CO₂ + 2H₂O + 125kJ
16g

2,5g CH₄ ----- 125 kj
16g CH₄ ------ X
X = (16×125)/2,5
X = 800 kJ = 800kJ per mole of methane

Answer 2

Answer: The enthalpy of combustion per mole of methane is 801.28 kJ.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of methane = 2.50 g

Molar mass of methane = 16 g/mol

Putting values in above equation, we get:

$\text{Moles of methane}=\frac{2.50g}{16g/mol}\\\\\text{Moles of methane}=0.156mol$

We are given:

Enthalpy of combustion for given mass = 125 kJ

Using unitary method, we get:

If 0.156 moles of methane has an enthalpy of combustion as 125 kJ.

So, 1 mole of methane has an enthalpy of combustion = $\frac{125kJ}{0.156mol}\times 1mol=801.28kJ$

Hence, the enthalpy of combustion per mole of methane is 801.28 kJ.