Kc for the decomposition of ammonium hydrogen sulfide is 1.8 x 10-4 at 15oC.

a. When 5.00 grams of the pure salt decomposes in a sealed 3.0 L flask at 15oC, what are the equilibrium concentrations of NH3 and H2S?
b. What are the equilibrium concentrations of the products at 15oC if 10.0 grams of the pure salt decomposes in the sealed flask?

.........NH4HS(s) ==> NH3(g) + H2S(g)

I........solid.........0........0
C........solid.........x........x
E........solid.........x........x

Kc = (NH3)(H2S)
1.8E-4 = x^2
x = 0.013M. Does this exhaust the 5.00 g NH4HS? 0.13 mols/L x 3L = 0.040 mols H2S and 0.040 mols NH3. How much NH4HS did we have initially. That's 5.00/51.112 = 0.0978 mol; therefore, there is some solid left after this decomposition.

That make the answer to the b part easier. Same answer as in part a. Because the solid is not part of the Kc expression.

To find the equilibrium concentrations of NH3 and H2S when 5.00 grams of ammonium hydrogen sulfide decomposes in a sealed 3.0 L flask at 15oC, we need to use the given equilibrium constant (Kc) and the stoichiometry of the reaction.

The balanced equation for the decomposition of ammonium hydrogen sulfide is:

NH4HS (s) ⇌ NH3 (g) + H2S (g)

First, we need to calculate the number of moles of NH4HS:

Number of moles of NH4HS = mass / molar mass
= 5.00 g / (14.01 g/mol + 1.01 g/mol + 32.06 g/mol)
= 5.00 g / 47.08 g/mol
≈ 0.106 mol

Since the initial volume of the flask is 3.0 L and the pressure is constant, the number of moles of NH3 and H2S at equilibrium will be the same. Let's denote the equilibrium concentration of NH3 and H2S as x.

Using the stoichiometry of the reaction, we can write the expression for the equilibrium concentration of NH3 as:

[NH3] = x

And for H2S:

[H2S] = x

The concentration of NH4HS can be calculated by subtracting the change from its initial concentration. Since the reaction is one-way, the change in concentration of NH4HS is -x.

[NH4HS] = [initial NH4HS] - [change in NH4HS]
= 0.106 mol / 3.0 L - (-x)
= 0.0353 M + x

Now, using the equilibrium constant expression:

Kc = ([NH3] * [H2S]) / [NH4HS]

We can substitute the equilibrium concentrations we calculated above:

1.8 x 10^(-4) = (x * x) / (0.0353 + x)

Simplifying this equation, we get a quadratic equation:

1.8 x 10^(-4) = x^2 / (0.0353 + x)

Rearranging and multiplying both sides by (0.0353 + x), we get:

(x^2) + (1.8 x 10^(-4))(0.0353 + x) = 0

Now we can solve this quadratic equation to find the value of x, which represents the equilibrium concentration of NH3 and H2S.

After solving this equation, we find that x is approximately 0.00530 M.

a. Therefore, the equilibrium concentrations of NH3 and H2S when 5.00 grams of the pure salt decomposes in a sealed 3.0 L flask at 15oC are approximately 0.00530 M.

To find the equilibrium concentrations of the products at 15oC if 10.0 grams of the pure salt decomposes in the sealed flask, we can follow the same steps described above.

Calculate the number of moles of NH4HS:

Number of moles of NH4HS = mass / molar mass
= 10.0 g / (14.01 g/mol + 1.01 g/mol + 32.06 g/mol)
= 10.0 g / 47.08 g/mol
≈ 0.213 mol

Since the initial volume of the flask is still 3.0 L, the number of moles of NH3 and H2S at equilibrium will be the same. Let's denote the equilibrium concentration of NH3 and H2S as x.

Using the stoichiometry of the reaction, we can write the expression for the equilibrium concentration of NH3 as:

[NH3] = x

And for H2S:

[H2S] = x

The concentration of NH4HS can be calculated by subtracting the change from its initial concentration. Since the reaction is one-way, the change in concentration of NH4HS is -x.

[NH4HS] = [initial NH4HS] - [change in NH4HS]
= 0.213 mol / 3.0 L - (-x)
= 0.0710 M + x

Using the equilibrium constant expression and substituting the equilibrium concentrations we calculated above, we can set up the equation:

Kc = ([NH3] * [H2S]) / [NH4HS]

1.8 x 10^(-4) = (x * x) / (0.0710 + x)

Solving this quadratic equation, we find that x is approximately 0.00771 M.

b. Therefore, the equilibrium concentrations of NH3 and H2S when 10.0 grams of the pure salt decomposes in a sealed 3.0 L flask at 15oC are approximately 0.00771 M.

To find the equilibrium concentrations of NH3 and H2S, we first need to write the balanced chemical equation for the decomposition of ammonium hydrogen sulfide (NH4HS):

NH4HS (s) ⇌ NH3 (g) + H2S (g)

The given value of Kc (equilibrium constant) tells us that at 15°C, the reaction is highly favored in the forward direction (formation of NH3 and H2S).

a. When 5.00 grams of the pure salt decomposes in a sealed 3.0 L flask at 15°C, we can calculate the number of moles of NH4HS:

Molar mass of NH4HS = (1 * 14.01 g/mol) + (1 * 1.01 g/mol) + (1 * 32.07 g/mol) = 63.09 g/mol

Moles of NH4HS = mass / molar mass = 5.00 g / 63.09 g/mol

Now, let's assume x moles of NH4HS decompose. This means x moles of NH3 and H2S are formed.

Since the initial moles of NH4HS are known (from the mass given), the equilibrium concentrations of NH3 and H2S can be expressed using the given volume of the flask (3.0 L):

[NH3] = x moles / 3.0 L
[H2S] = x moles / 3.0 L

To find x, we can use the equation for Kc:

Kc = [NH3] * [H2S] / [NH4HS]
1.8 x 10^-4 = (x / 3.0) * (x / 3.0) / (5.00 / 63.09)

Simplifying the equation gives:

(x^2) / 9.0 = (1.8 x 10^-4) * (5.00 / 63.09)

Solving for x, we find:

x^2 = (9.0) * (1.8 x 10^-4) * (5.00 / 63.09)
x = √((9.0) * (1.8 x 10^-4) * (5.00 / 63.09))

Once we have the value of x, we can substitute it back into [NH3] and [H2S]:

[NH3] = x moles / 3.0 L
[H2S] = x moles / 3.0 L

b. To find the equilibrium concentrations when 10.0 grams of the pure salt decompose in the sealed flask, the process is similar. Follow the same steps as in part a, using the new value of initial moles of NH4HS (calculated from the new mass given) and solve for x. Finally, substitute x into [NH3] and [H2S] expressions to find the equilibrium concentrations.