Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13
Answer:- 3.12 g carbon tetrachloride are needed.
Solution:- The balanced equation is:

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.
percent yield formula is:
percent yield = 


theoretical = 3.44 g
From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.
Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

= 
So, 3.12 grams of carbon tetrachloride are needed to be reacted.
Of course we would experience them.