The transformation occurring in Duma's method is given below
$$C_2H_7N + 2x + \frac{y}{2}CuO \rightarrow xCO_2 + \frac{y}{2}H_2O + \frac{z}{2}N_2 + 2x + \frac{y}{2}Cu$$
The value of y is _________. (Integer answer)

We start with the un-balanced skeletal equation given for Duma’s method:

$$C_2H_7N +\left(2x+\dfrac{y}{2}\right)CuO \;\longrightarrow\; x\,CO_2 +\dfrac{y}{2}H_2O +\dfrac{z}{2}N_2 +\left(2x+\dfrac{y}{2}\right)Cu$$

To obtain the numerical values of the unknown coefficients, we now equate the number of atoms of each element on the two sides of the arrow.

Carbon balance
There are $$2$$ carbon atoms in one molecule of $$C_2H_7N$$ on the left. On the right, every $$CO_2$$ molecule contains one carbon atom, and the coefficient in front of $$CO_2$$ is $$x$$. So we write

$$x = 2$$

Hydrogen balance
On the left, the compound $$C_2H_7N$$ provides $$7$$ hydrogen atoms. On the right, each molecule of $$H_2O$$ has two hydrogens and its coefficient is $$\dfrac{y}{2}$$, giving a total of

$$\left(\dfrac{y}{2}\right)\times 2 = y$$ hydrogen atoms.

Equating the two sides, we have

$$y = 7$$

Nitrogen balance
The left side supplies one nitrogen atom. On the right, a molecule of $$N_2$$ contains two nitrogen atoms and its coefficient is $$\dfrac{z}{2}$$, giving

$$\left(\dfrac{z}{2}\right)\times 2 = z$$ nitrogen atoms. Hence

$$z = 1$$

Oxygen balance (consistency check)
The coefficient in front of $$CuO$$ is $$2x + \dfrac{y}{2}$$, so the number of oxygen atoms on the left is also $$2x + \dfrac{y}{2}$$. On the right, oxygen appears in $$CO_2$$ and $$H_2O$$. Counting them gives

$$\underbrace{x\times 2}_{\text{from } CO_2} + \underbrace{\left(\dfrac{y}{2}\right)\times 1}_{\text{from } H_2O} = 2x + \dfrac{y}{2}$$

Substituting $$x = 2$$ and $$y = 7$$, both sides give $$2(2) + \dfrac{7}{2} = 4 + \dfrac{7}{2}$$, confirming that oxygen is already balanced.

Thus, the integer value required for $$y$$ is

$$7$$

Hence, the correct answer is Option 7.