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Mama L [17]
3 years ago
10

When hydrochloric acid is poured over potassium sulfide, 43.7 mLmL of hydrogen sulfide gas is produced at a pressure of 758 torr

torr and 26.0 ∘C?
Chemistry
1 answer:
kvasek [131]3 years ago
6 0

Answer:

0.196 grams of K2S reacted

Explanation:

When hydrochloric acid is poured over potassium sulfide, 43.7 mLmL of hydrogen sulfide gas is produced at a pressure of 758 torrtorr and 26.0 ∘C

How much potassium sulfide has reacted in grams?

Step 1: Data given

Volume of hydrogen sulfide (H2S) produced = 43.7 mL

Pressure = 758 torr = 758/760 = 0.9973684 atm

Temperature = 26.0 °C = 273 + 26 = 299 K

Step 2: The balanced equation

2 HCl + K2S → H2S + 2 KCl

Step 3: Calculate moles H2S

p*V = nRT

n = (pV)/(RT)

⇒ with n= the number of moles of H2S

⇒ with p = the pressure = 0.9973684 atm

⇒ with V = the volume of the gas = 43.7 mL = 0.0437 L

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

⇒ with T = The temperature = 26°C = 299 Kelvin

n = (0.9973684 * 0.0437)/ (0.08206*299)

n = 0.001776 moles H2S

Step 4: calculate moles of K2S

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

For 0.001776 moles H2S produced, we need 0.001776 moles K2S

Step 5: Calculate mass of K2S

Mass K2S = moles K2S * molar mass K2S

Mass K2S = 0.001776 moles * 110.26 g/mol

Mass K2S = 0.196 grams K2S

0.196 grams of K2S reacted

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3 0
4 years ago
Gold cylinder has a mass of 75 g and a specific heat of 0.129J/G degrees Celsius it is heated to 65°C and then put in 500 g of w
nadezda [96]
<h3>Answer:</h3>

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<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of gold cylinder as 75 g
  • specific heat of gold is 0.129 J/g°C
  • Initial temperature of gold cylinder is 65°C
  • Mass of water is 500 g
  • Initial temperature of water is 90 °C

We are required to calculate the final temperature;

  • We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
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<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>

Assuming the final temperature is X° C

Then; ΔT = (X-65)°C

Therefore;

Q = 75 g × 0.129 J/g°C × (X-65)°C

   = 9.675X - 628.875 Joules

<h3>Step 2: Calculate the quantity of heat released by water</h3>

Taking the final temperature as X° C

Change in temperature, ΔT = (90 - X)° C

Specific heat capacity of water is 4.184 J/g°C

Therefore;

Q = 500 g × 4.184 J/g°C × (90 - X)° C

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<h3>Step 3: Calculate the final temperature, X°C</h3>

we know that the heat gained by gold cylinder is equal to the heat released by water.

9.675X - 628.875 Joules = 188,280 -2092X joules

2101.675 X = 188908.875

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