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galben [10]
3 years ago
14

State whether the following statements are true or false. If false, explain why. (a) A reaction stops when equilibrium is reache

d. (b) An equilibrium reaction is not affected by increasing the concentrations of products. (c) If one starts with a higher pressure of reactant, the equilibrium constant will be larger. (d) If one starts with higher concentrations of reactants, the equilibrium concentrations of the products will be larger.
Chemistry
1 answer:
Elden [556K]3 years ago
7 0

Answer:

(a) False;

(b) False;

(c) False;

(d) True.

Explanation:

(a) When equilibrium is reached, the forward reaction rate becomes equal to the reverse reaction rate, that's why the molarity of each species remains constant, but reactions don't stop.

(b) According to the principle of Le Chatelier, an increase in molarity of either reactants or products would lead to a disturbance of equilibrium. This disturbance would lead to the shift of equilibrium towards the side which would minimize such a disturbance.

(c) Equilibrium constant is only temperature-dependent, it's independent of molarity, pressure, volume etc. of any species present in the reaction.

(d) The greater the initial molarity of reactants, the more products can be formed, e. g., since the ratio of products to reactants should be kept constant, the larger the amount of reactants, the greater the amount of products formed to keep a constant ratio.

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Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

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Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

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67.8 / 100 = 30.25 / Theoretical yield

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Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

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Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

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Molar mass of AlI₃ = 27 + (3×127)

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Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

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From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

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