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2 years ago

What is 39483048^349374*3948048/3i4u4

Engineering

1 answer:

Verizon [17]2 years ago

3 0

Answer:

1. $3.81813506×10^2^9$

2. $1.71479428×10^6^5$

3. $9.38483383×10^2^6$

4. $1.150847×10^2^9$

Explanation:

Feel free to give brainliest

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Read 2 more answers

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

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Answer:

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Explanation:

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