After one side of the stock is planed to a true surface, it is usually

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>

Learn more about theoretical and actual yield here:

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7 0

3 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

A city is experiencing a windstorm. The wind has blown away some of the houses in that city. What load bearing factor did the ar

yuradex [85]

Answer:

oa

Explanation:

6 0

3 years ago

When was the first black or white hole discovered?

ivolga24 [154]

Answer:

1964

Explanation:

It was discovered in 1964 when a pair of Geiger counters were carried on board a sub-orbital rocket launched from New Mexico.

3 0

3 years ago

What is the proper distance from the feet of a straight ladder to the wall?

blondinia [14]

Answer:

One Foot

Explanation:

The general rule is to maintain a distance of one foot per four feet of height.

3 0

3 years ago