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We have a mixture consisting of 90 mole% (45 mol/liter) and 10 mole % impurity B (5 mol/liter). To be of satisfying quality the mole ratio of A to B in the mixture must be 100 to 1 or higher. D reacts with both A and B as follows: A + D → R (i) B+ D + S (ii) (a) Estimate the concentrations of A and B such that the ratio of concentrations of A to B is 100. (b) Calculate the concentration of D that must be added at the beginning of the reaction. Data: -TA = 21CACD -1B = 147CBCD



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IYesntMath

Answer:

In this problem, we need to determine the concentrations of A and B to achieve a mole ratio of 100:1 and calculate the initial concentration of D required for the reaction.

Given:

Mixture consists of 90 mole% A (45 mol/liter) and 10 mole% B (5 mol/liter)

Desired mole ratio of A to B is 100:1

Reactions: A + D → R and B + D → S

Reaction rates: -rA = 21CACD and -rB = 147CBCD

Step 1: Determine the concentrations of A and B to achieve a mole ratio of 100:1.

Let the concentration of B be x mol/liter.

Then, the concentration of A will be 100x mol/liter.

Total concentration = Concentration of A + Concentration of B

50 mol/liter = 100x + x

50 mol/liter = 101x

x = 50/101 ≈ 0.495 mol/liter

Concentration of A = 100 × 0.495 ≈ 49.5 mol/liter

Concentration of B ≈ 0.495 mol/liter

Step 2: Calculate the initial concentration of D required for the reaction.

We can use the reaction rate equations to determine the concentration of D.

-rA = 21CACD

-rB = 147CBCD

Since both reactions consume D, we can equate the reaction rates:

-rA = -rB

21CACD = 147CBCD

Substitute the concentrations of A and B:

21 × 49.5 × CD = 147 × 0.495 × CD

1039.5CD = 72.765CD

CD = 72.765/1039.5 ≈ 0.07 mol/liter

Therefore, the initial concentration of D should be approximately 0.07 mol/liter.

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