Question

A mixture of N₂ and H₂ has initially mass ratio of 196 : 1 then find after how many steps we can obtain a mixture containing 1 : 14 mole ratio of N₂ and H₂.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

Answer 2

The initial mass ratio of N₂ to H₂ is 196:1. The molar mass of N₂ is approximately 28 g/mol. The molar mass of H₂ is approximately 2 g/mol.

Let the mass of N₂ be $196x$ and the mass of H₂ be $1x$. The initial number of moles of N₂ is: $n_{N_2, \text{initial}} = \frac{196x}{28} = 7x$

The initial number of moles of H₂ is: $n_{H_2, \text{initial}} = \frac{1x}{2} = 0.5x$

The initial mole ratio of N₂ to H₂ is: $\frac{n_{N_2, \text{initial}}}{n_{H_2, \text{initial}}} = \frac{7x}{0.5x} = 14:1$

The target mole ratio of N₂ to H₂ is 1:14. $\frac{n_{N_2, \text{final}}}{n_{H_2, \text{final}}} = \frac{1}{14}$

The problem asks for the number of "steps" to achieve this change. The term "steps" is not explicitly defined in the context of a chemical reaction. However, to change the mole ratio from N₂-rich (14:1) to H₂-rich (1:14), the relative amount of N₂ must decrease significantly compared to H₂.

The most straightforward interpretation, given the lack of defined reaction conditions or process steps, is to consider a single operation that achieves the target ratio. This could be a process of selectively removing N₂ from the mixture.

If we assume the amount of H₂ remains constant ($n_{H_2, \text{final}} = n_{H_2, \text{initial}} = 0.5x$), we can find the required final amount of N₂: $\frac{n_{N_2, \text{final}}}{0.5x} = \frac{1}{14}$ $n_{N_2, \text{final}} = \frac{0.5x}{14} = \frac{x}{28}$

This means that the amount of N₂ needs to be reduced from $7x$ to $x/28$. The amount of N₂ to be removed is $7x - \frac{x}{28} = \frac{196x - x}{28} = \frac{195x}{28}$.

If this selective removal of N₂ to reach the target ratio is considered a single "step" or operation, then the answer is 1. Standard chemical reactions like the Haber-Bosch process (N₂ + 3H₂ → 2NH₃) would consume H₂ faster than N₂, thus increasing the N₂:H₂ ratio, moving away from the target if N₂ is initially in excess. Therefore, a process involving selective removal or addition is implied. Assuming the simplest interpretation of "steps" as a single process to achieve the desired ratio, the answer is 1.