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Ronch [10]
3 years ago
15

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) You find the reaction to be first

order with respect to HBr and first order with respect to O2. You propose the following mechanism: HBr(g) + O2(g) → HOOBr(g) HOOBr(g) + HBr(g) → 2 HOBr(g) HOBr(g) + HBr(g) → H2O(g) + Br2(g) a. Confirm that the elementary reactions add to give the overall reaction. (Hint: Use Hess Law) b. Based on the experimentally determined rate law, which step is rate determining? c. What are the intermediates in this mechanism? d. If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

Chemistry
1 answer:
Vlada [557]3 years ago
4 0

Answer:

Explanation:

FIND THE SOLUTION IN THE ATTACHMENT

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yield = 52.23 %

Explanation:

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mass of aluminium = mass of bottle with aluminium pieces - bottle mass

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mass of alum = mass of bottle with final product - bottle mass

mass of alum = 19.0414 - 9.8981 = 9.1433 g

number of moles = mass / molecular weight

number of moles of aluminium = 0.9974 / 27 = 0.03694 moles

number of moles of alum (practical) = 9.1433 / 474 = 0.01929 moles

To calculate the theoretical quantity of alum that should be obtained from 0.03694 moles of aluminium we devise the following reasoning:

if       2 moles of aluminium produce 2 moles of alum

then 0.03694 moles of aluminium produce X moles of alum

X = (0.03694 × 2) / 2 = 0.03694 moles of alum (theoretical)

yield = (practical quantity / theoretical quantity) × 100

yield = (0.01929 /  0.03694) × 100

yield = 52.23 %

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reaction yield

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5 0
4 years ago
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astra-53 [7]

Answer:

C.

Explanation:

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6 0
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Answer:

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antiseptic1488 [7]

Answer:

aluminium

Explanation:

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