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Calorimetry can be used to find heats of combustion. suppose a calorimeter helps someone to determine that 4032 kj of heat is released during the combustion of 88.0 grams of propane, c3h8 (44.0 g/mol). what would be the amount of heat released for the combustion of 0.500 moles of propane?

Respuesta :

superman1987

Answer:

1008.0 kJ.

Explanation:

  • Firstly, we need to calculate the no. of moles of 88.0 g of propane:

n = mass/molar mass = (88.0 g)(/(44.0 g/mol) = 2.0 mol.

∴ The combustion of 2.0 moles of propane produces 4032.0 kJ.

Using cross multiplication:

The combustion of 2.0 moles of propane produces → 4032.0 kJ.

The combustion of 0.5 moles of propane produces → ??? kJ.

∴ The amount of heat released for the combustion of 0.5 moles of propane = (4032.0 kJ)(0.5 mol)/(2.0 mol) = 1008.0 kJ.

Answer:

1,075.5 kiloJoules of heat released for the combustion of 0.500 moles of propane.

Explanation:

Heat released on combustion of 88.0 grams of propane = 4,032 kJ

Moles of propane = [tex]\frac{88.0 g}{44.0 g/mol}=2 mol[/tex]

Heat released on combustion of 2 moles of propane = 4,032 kJ

Heat released on combustion of 0.500 moles of propane :

[tex]\frac{4,302 kJ}{2}\times 0.500=1,075.5 kJ[/tex]

1,075.5 kiloJoules of heat released for the combustion of 0.500 moles of propane.

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