The transformation occurring in Duma's method is given below : $$\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N} +\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{CuO} \rightarrow \mathrm{xCO}_{2}+\frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O}+\frac{\mathrm{z}}{2} \mathrm{~N}_{2}+\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{Cu}$$ The value of $y$ is ____. (Integer answer)

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Practical and Analytical Chemistry

The transformation occurring in Duma's method is given below :

C_{2} H_{7} N + (2 x + \frac{y}{2}) CuO \to xCO_{2} + \frac{y}{2} H_{2} O + \frac{z}{2} N_{2} + (2 x + \frac{y}{2}) Cu

The value of

y

is ____. (Integer answer)

Answer: 7

Solution

Balancing atoms in the Dumas combustion of

C_{2} H_{7} N

: carbon gives

x = 2

and nitrogen gives

z = 1

. Every hydrogen leaves as water, and the coefficient of water is

y /2

, so

y

hydrogen atoms appear as

\frac{y}{2} H_{2} O

; matching the seven hydrogens of

C_{2} H_{7} N

gives

y = 7

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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