Question

A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3. Once the reaction reaches equilibrium, what mass of NH3 is present in the container

Answer 1

Answer:

The mass of $NH_{3}$ in the container is 2.074 gram

Explanation:

Given:

Volume of $NH_{4} HS$    $V = 2$ lit

Equilibrium constant $k _{eq} = 3.5 \times 10^{-3}$

The reaction in which $NH_{3}$ is produced

  $NH_{4} HS$ ⇄ $NH_{3} + H_{2}S$

Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

From the formula of equilibrium constant,

  $k_{eq} = (NH_{3})(H_{2}S )$

   $x^{2} = 3.5 \times 10^{-3}$

     $x = 0.061$ M

Above value shows,

   $NH_{3} = 0.061$ $\frac{moles}{L}$

So in 2 L no. moles of $NH_{3}$ = $0.061 \times 2 = 0.122$ moles.

So mass of 0.122 mole of $NH_{3}$ is = $0.122 \times 17 = 2.074$ g

Therefore, the mass of $NH_{3}$ in the container is 2.074 gram