Let's solve this step-by-step.

Given:

• Mass of propane combusted = 2.00 g
• Enthalpy change for combustion of 2.00 g propane = −102 kJ
• We want: Enthalpy change for combustion of 1 mole of propane (C₃H₈)

Step 1: Calculate the molar mass of propane (C₃H₈)

• Carbon (C) atomic mass ≈ 12.01 g/mol
• Hydrogen (H) atomic mass ≈ 1.008 g/mol

Molar mass of C₃H₈ = (3 × 12.01) + (8 × 1.008)
= 36.03 + 8.064
= 44.094 g/mol

Step 2: Calculate moles of propane in 2.00 g

moles=massmolar mass=2.00 g44.094 g/mol≈0.0454 mol\text{moles}=\frac{\text{mass}}{\text{molar mass}}=\frac{2.00\text{ g}}{44.094\text{ g/mol}}\approx 0.0454\text{ mol}moles=molar massmass​=44.094 g/mol2.00 g​≈0.0454 mol

Step 3: Calculate enthalpy change per mole

Given that −102 kJ is for 0.0454 mol, then for 1 mole:

ΔH=−102 kJ0.0454 mol≈−2248 kJ/mol\Delta H=\frac{-102\text{ kJ}}{0.0454\text{ mol}}\approx -2248\text{ kJ/mol}ΔH=0.0454 mol−102 kJ​≈−2248 kJ/mol

Final answer:

The enthalpy change for the combustion of 1 mole of propane under the same conditions is approximately −2248 kJ/mol. If you want, I can also help you compare this with standard enthalpy values or explain the combustion reaction involved!