22) C

Number of Cr on LHS = 2
Number of Cr on RHS = d
So d = 2 (there are no other Cr on the left and right hand sides)

Total charge on left side so far = -2 + 1 = -1
Total charge on right side = 2 x (+3) = +6
To balance charges, b = +6 – (-2) = + 6 + 2 = +8
b = 8

Let us form an equation for the total number of H

4a + b = 4c + 2e
4a + b – 4c = 2e
e = 4a + b – 4c
             2

Let us form an equation for the total number of O

a + 7 = 2c + e
e = a + 7 – 2c

Putting the two equations together,

4a + b – 4c  =  a + 7 – 2c
2

4a + b – 4c = 2 (a + 7 – 2c)
4a + b – 4c = 2a + 14 – 4c
4a + b = 2a + 14
2a + b = 14

Substituting value of b,
2a + 8 = 14
2a = 6
a = 3

Number of carbons on LHS = 2 x a = 2 x 3 = 6
Number of carbons on RHS = 2c
2c must equal to 6
c = 3

Using the equation (e = a + 7 – 2c),
e = 3 + 7 – 2(3)
e = 10 – 6
e = 4