ExampleCO (g) + NO(g) --> CO2(g) + (1/2)N2(g)

• DH = -373.3 kJ

• If 27.3 g of CO are present, what is the enthalpy change?

• 27.3g CO = 0.975 mol CO

• 0.975 mol CO x -373.3 kJ/mol CO

• -364 kJ

Remember, since DH is negative, enthalpy is a product.

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