Answer
$P_{NH_3} = 0.33 \space atm$
Work Step by Step
1. Write the $K_p$ expression:
$$K_p = P_{H_2S} P_{NH_3}$$
2. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& P_{ H_2S }& P_{ NH_3 }\\
Initial& 0 & 0 \\
Change& +x& +x\\
Equilibrium& 0 +x& 0 +x\\
\end{vmatrix}$$
3. Substitute and solve for x:
$$0.11 = (x)(x)$$ $$0.11 = x^2$$ $$x = \sqrt{0.11} = 0.33 \space atm$$
$P_{NH_3} = x = 0.33 \space atm$
4. Determine if its possible to produce this pressure with 55.0 g of reactant.
$$Temperature/K = 250 + 273.15 = 520 \space K$$
5. According to the Ideal Gas Law:
$$n = \frac{PV}{RT} = \frac{( 0.33 \space atm )( 5.0 \space L )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 520 \space K)}$$
$$n = 0.039 \space mol \space NH_3$$
6. Calculate or find the molar mass for $ NH_4HS $:
$ NH_4HS $ : ( 1.008 $\times$ 5 )+ ( 14.01 $\times$ 1 )+ ( 32.07 $\times$ 1 )= 51.12 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 55.0 \space g \space NH_4HS \times \frac{1 \space mole}{ 51.12 \space g} = 1.08 \space moles \space NH_4HS$$
Yes, 1.08 mol of $NH_4HS$ can produce $0.039$ mol $NH_3$, so the answer is correct, 0.33 atm.
