Question

Using the following equation for the combustion of octane, calculate the heat of reaction for 75.00 g of octane. The molar mass of octane is 114.33 g/mole.2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ

Answer 1

1) Write the chemical equation

`2C_8H_(18)+25O_2\rightarrow12CO_2+18H_2O`
`\Delta Hº_(rxn)=-11018kJ`

2) Convert grams of octane to moles of octane

The molar mass of octane is 114.33 g/mol

`_{}molC_8H_(18)=75.00gC_8H_(18)\cdot(1molC_8H_(18))/(114.33gC_8H_(18))=0.6560molC_8H_(18)`

3) Calculate the heat of the reaction

Use the heat of reaction of the balanced reaction to make a conversion factor.

2 C8H18 = -11018kJ

`\Delta H=0.6560molC_8H_(18)\cdot(-11018kJ)/(2molC_8H_(18))=-3613.9kJ`

The heat of reaction for 75.00 g of octane is -3614kJ.

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