we would need 0.037 mol. We know that because of a formula we can use and some data from the problem. The formula would be:
n= Number of heat in kj x 2 moles
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5460 kJ
The number of Kj of heat is provided(100 kJ). The standard enthalpy of combustion is -5460.0 kj. so we multiply 100 kj and 2 moles and then we divide on 5460 kj and the answer is 0.037 moles.