Q20 / 150 · MHT CET 2026 (15 April Shift 2) · Chemistry
Find $\Delta n$ when 1 mol of each $\text{NH}_{3(g)}$ and $\text{HCl}_{(g)}$, reacts to form solid $\text{NH}_4\text{Cl}$.
- A$1$
- Correct answerB$-2$
- C$-1$
- D$-3$
Solution
The balanced chemical equation for the reaction is:
$\text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)}$
The change in the number of moles of gaseous species, $\Delta n$, is given by the difference between the number of moles of gaseous products and gaseous reactants.
Number of moles of gaseous products, $n_P = 0$
Number of moles of gaseous reactants, $n_R = 1 + 1 = 2$
$\Delta n = n_P - n_R = 0 - 2 = -2$
Answer: $-2$
$\text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)}$
The change in the number of moles of gaseous species, $\Delta n$, is given by the difference between the number of moles of gaseous products and gaseous reactants.
Number of moles of gaseous products, $n_P = 0$
Number of moles of gaseous reactants, $n_R = 1 + 1 = 2$
$\Delta n = n_P - n_R = 0 - 2 = -2$
Answer: $-2$
Hence, the correct answer is option: 'B'.