Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 4.00 mol of al undergoes the reaction with a stoichiometrically equivalent amount of fe2o3.

Given:

Enthalpy change ΔH for the thermite reaction = -850 kJ

Moles of Al involved = 4

To determine:

Reaction enthalpy when 4 moles of Al reacts

Explanation:

The thermite reaction is-

2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)

When 2 moles of Al react the enthalpy change is -850 kJ

therefore, when 4 moles of Al reacts, the change in enthalpy is-

= 4 moles * (-850) kJ/2 moles = -1700 kJ

Ans: Enthalpy change is -1700 kJ

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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am i too late?

explanation: