Question

When hydrochloric acid is poured over potassium sulfide, 42.3 mL of hydrogen sulfide gas is produced at a pressure of 750 torr and 25.6 ∘C.

Part A Write an equation for the gas-evolution reaction. 2HCl+KS(s)→H2S(g)+KCl(aq) HCl+K2S(s)→H2S(g)+KCl(aq) HCl+KS(s)→H2S(g)+KCl(aq) 2HCl+K2S(s)→H2S(g)+2KCl(aq)
Part B Determine how much potassium sulfide (in grams) reacted.

Answer 1

2 HCl(aq) + K2S (s) → H2S(g)+2KCl(aq)

There will react 0.187 grams of K2S

Step-by-step explanation:

Step 1: Data given

Volume of H2S produced = 42.3 mL

Pressure = 750 torr

Temperature = 25.6 °C

Step 2: The balanced equation

2 HCl(aq) + K2S (s) → H2S(g)+2KCl(aq)

Step 3: Calculate moles of H2S

P*V = n*R*T

⇒ with P = the pressure of the H2S gas = 750 torr = 750/760 atm

⇒ with V = the volume of H2S = 0.0423 L

⇒ with n = the number of moles of H2S = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25.6 °C = 298.75 Kelvin

n = (P*V)/(R*T)

n(H2S) = 0.0017 moles

Step 4: Calculate moles of K2S

For 2 moles of HCl we need 1 mol of K2S to produce 1 mol of H2S and 2 moles of KCl

For 0.0017 moles H2S produced, we have 0.0017 moles of K2S

Step 5: Calculate mass of K2S

Mass K2S = moles K2S * molar mass K2S

Mass K2S = 0.00170 moles * 110.26 g/mol

Mass K2S = 0.187 grams K2S

There will react 0.187 grams of K2S

Answer 2

Part. A

2HCl+K2S(s) → H2S(g)+2KCl(aq)

Part B.

0.19 g of K2S reacted

Step-by-step explanation:

Let's use the Ideal Gas Law Equation to know the moles of H2S formed under those conditions.

We have to do this conversion before

750 Torr = 0.98 atm

25.6 °C = 25.6°C + 273 > 298.6K

42.3 mL = 0.0423L

Pressure. Volume = n . R . T

Let's replace the values

0.98 atm. 0.0432L = n . 0.082 L.atm/m.K . 298.6K

(0.98 atm. 0.0432L) / (0.082 m.K/L.atm . 298.6K) = n

1.7x10*-3 = n

So now let's verify the reaction.

Ratio is 1:1

1 mol of K2S makes 1 mol of H2S

Molar mass K2S = 110.26 g/m

1.7x10*-3 moles . 110.26 g/m = 0.19 g