18) D

Let us form an equation for the number of O atoms present in the reaction equation.

eq 1: **3b = 6a + c + 2**

Let us form an equation for the number of H atoms present in the reaction equation.

eq 2: **b = 2c**

Let us form an equation for the number of N atoms present in the reaction equation.

eq 3: **b = 2a + 2**

Using the second and third equations,

b = 2c = 2a + 2

2c = 2a + 2

2(c) = 2(a + 1)

c = a + 1

Submitting this expression of ‘c’ into the second equation,

b = 2c

b = 2 (a + 1)

b = 2a + 2

3b = 3 (2a + 2) = 6a + 6

Substituting the expression for 3b into the first equation,

3b = 6a + c + 2

6a + 6 = 6a + c + 2

6 = c + 2

c = 4

Substituting the value of ‘c’ into the second equation

b = 2c

b = 2 (4)

b = 8