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ololo11 [35]
2 years ago
11

The equilibrium constant, Kc, for the following reaction is 3.61×10-4 at 426 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar

ge sample of PCl5(g) is introduced into an evacuated vessel at 426 K, the equilibrium concentration of Cl2(g) is found to be 0.327 M. Calculate the concentration of PCl5 in the equilibrium mixture.

Chemistry
1 answer:
iogann1982 [59]2 years ago
8 0

Answer:

the concentration of PCl5 in the equilibrium mixture = 296.20M

Explanation:

The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.

The detaied steps is as shown in the attached file.

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Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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