The nuclear equation for the fusion reaction between deuterium and tritium is:

2H + 3H -> 4He + 1n

The mass of 1 mole of deuterium is 2 g/mol, which is equivalent to 2/Avogadro's number = 2/6.02214 x 10^23 = 3.323 gram.

To calculate the energy released in MeV, we use Einstein's mass-energy equation:

E = Δm x c^2

where E is the energy, Δm is the change in mass, and c is the speed of light.

First, we need to calculate the change in mass.
The mass of 1 mole of deuterium is 3.323 gram, and the mass of helium-4 is 4.001506 gram/mol. The change in mass (Δm) is the difference between these two masses:

Δm = 4.001506 g/mol - 3.323 g/mol = 0.678506 g/mol

Next, we convert this mass to kilograms:

Δm = 0.678506 g/mol x (1 kg / 1000 g) = 0.000678506 kg/mol

Putting these values into the equation for energy:

E = Δm x c^2 = 0.000678506 kg/mol x (3 x 10^8 m/s)^2 = 0.000678506 kg/mol x (9 x 10^16 m^2/s^2) = 6.106554 x 10^13 Joules/mol

Lastly, we convert this energy to MeV (mega-electron volts).
1 Joule = 6.242 x 10^18 MeV, so:

E = 6.106554 x 10^13 J/mol x (6.242 x 10^18 MeV/J) = 3.811946 x 10^32 MeV/mol

Therefore, the energy released when 1 mole of deuterium reacts completely is approximately 3.811946 x 10^32 MeV.

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