I shall assume the following equations are the ones of interest as none was provided:
(a) 63 Li + 56 28Ni →?
(b) 40 20Ca + 248 96Cm→147 62Sm + ?
(c) 88 38Sr + 84 36Kr→116 46Pd + ?
(d) 40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?
I shall also assume in the following calculations that the mass number of the unknown products is 'a' and the atomic number of the unknown products is 'b'.
In balancing nuclear equations, we observe the following rules:
I. The total number of protons and neutrons in the products and reactants must be the same ( conservation of mass number, A ).
II. The total number of nuclear charges in the products and reactants must be the same ( conservation of atomic number, Z ).
For (a),
63 Li + 56 28Ni→?
Add mass number, A, of reactants = 6+56 = 62
Add atomic number, Z, of reactants = 3+28 = 31
A of reactants = A of unknown product = 62
and, Z of reactants = Z of unknown product = 31
Hence, unknown product is the element with Z as 31 = Ga
Balanced equation becomes:
63 Li + 56 28Ni → 62 31Ga
For (b),
40 20Ca + 248 96Cm→147 62Sm + ?
A of reactants = 40+248 = 288
A of products = 147 + 'a'
But A of reactants = A of products
288 = 147 + 'a'
‘a’ = 288 – 147 = 141
Also,
Z of reactants = 20 + 96 = 116
Z of products = 62 + 'b'
Z of reactants = Z of products
116 = 62 + 'b'
'b' = 116 – 62 = 54
So, unknown product is the element with Z = 54 = Xe
Balanced equation is:
40 20Ca + 248 96Cm→147 62Sm + 141 54Xe
For (c),
88 38Sr + 84 36Kr→116 46Pd + ?
A of reactants = 88 + 84 = 172
A of products = 116 + 'a'
But A of reactants = A of products
172 = 116 + 'a'
‘a’ = 172 – 116 = 56
Also,
Z of reactants = 38 + 36 = 74
Z of products = 46 + 'b'
Z of reactants = Z of products
74 = 46 + 'b'
'b' = 74 – 46 = 28
So, unknown product is the element with Z = 28 = Ni
Balanced equation is:
88 38Sr + 84 36Kr→116 46Pd + 56 28Ni
For (d),
40 20Ca + 238 92U→70 30Zn + 4 10 n + 2?
A of reactants = 40 + 238 = 278
A of products = 70 + (4 x 1) + (2 x 'a') = 70 + 4 + 2a = 74 + 2a
But, A of reactants = A of products
278 = 74 + 2a
2a = 278 – 74
2a = 204
'a' = 204/2 = 102
Also,
Z of reactants = 20 + 92 = 112
Z of products = 30 + (4 x 0) + (2 x b) = 30 + 0 + 2b = 30 + 2b
Z of reactants = Z of products
112 = 30 + 2b
2b = 112 – 30
2b = 82
‘b’ = 82/2 = 41
Therefore, unknown product is element with Z = 41 =Nb
Balanced equation is:
40 20Ca + 238 92U→70 30Zn + 4 10 n + 2 102 41Nb
