Answer:
The mass of [tex]NH_{3}[/tex] in the container is 2.074 gram
Explanation:
Given:
Volume of [tex]NH_{4} HS[/tex] [tex]V = 2[/tex] lit
Equilibrium constant [tex]k _{eq} = 3.5 \times 10^{-3}[/tex]
The reaction in which [tex]NH_{3}[/tex] is produced
[tex]NH_{4} HS[/tex] ⇄ [tex]NH_{3} + H_{2}S[/tex]
Here equal moles of [tex]NH_{3}[/tex] and [tex]H_{2}S[/tex] is formed.
From the formula of equilibrium constant,
[tex]k_{eq} = (NH_{3})(H_{2}S )[/tex]
[tex]x^{2} = 3.5 \times 10^{-3}[/tex]
[tex]x = 0.061[/tex] M
Above value shows,
[tex]NH_{3} = 0.061[/tex] [tex]\frac{moles}{L}[/tex]
So in 2 L no. moles of [tex]NH_{3}[/tex] = [tex]0.061 \times 2 = 0.122[/tex] moles.
So mass of 0.122 mole of [tex]NH_{3}[/tex] is = [tex]0.122 \times 17 = 2.074[/tex] g
Therefore, the mass of [tex]NH_{3}[/tex] in the container is 2.074 gram
