Consider the following reaction sequence The percentage of nitrogen in product ‘T’ formed is [2026]

Question

Consider the following reaction sequence

The percentage of nitrogen in product ‘T’ formed is ______ %. (Nearest integer)
(Given molar mass in g ${mol}^{- 1}$ : H : 1, C : 12, N : 14, O : 16) [2026]

Smart Achievers · Accepted Answer

(20)

$Mol. wt = 6 \times 12 + (6 \times 1) + (2 \times 14) + (2 \times 16)$ $= 138$

$% N = \frac{28}{138} \times 100 = 20.29 %$

Solution

Q.
Consider the following reaction sequence

The percentage of nitrogen in product ‘T’ formed is ______ %. (Nearest integer)
(Given molar mass in g ${mol}^{- 1}$ : H : 1, C : 12, N : 14, O : 16) [2026]

Ans.
(20)

$Mol. wt = 6 \times 12 + (6 \times 1) + (2 \times 14) + (2 \times 16)$ $= 138$

$% N = \frac{28}{138} \times 100 = 20.29 %$

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Solution

Q. Consider the following reaction sequence The percentage of nitrogen in product ‘T’ formed is ______ %. (Nearest integer) (Given molar mass in g mol-1: H : 1, C : 12, N : 14, O : 16) [2026]

Ans. (20) Mol. wt=6×12+(6×1)+(2×14)+(2×16)=138 %N=28138×100=20.29%

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Q.
Consider the following reaction sequence

The percentage of nitrogen in product ‘T’ formed is ______ %. (Nearest integer)
(Given molar mass in g ${mol}^{- 1}$ : H : 1, C : 12, N : 14, O : 16) [2026]

Ans.
(20)

$Mol. wt = 6 \times 12 + (6 \times 1) + (2 \times 14) + (2 \times 16)$ $= 138$

$% N = \frac{28}{138} \times 100 = 20.29 %$