Someone can help me with the question 3, please

Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

5 0

3 years ago

If a 3 kg moon rock is moved from the moon to the earth does the mass of the rock change

sukhopar [10]

Answer:

No, only the wiehgt, because the gravitational pull on the two, are diffrent. 3kg on the moon may be 6kg here on earth. but in the end, the rock still holds however much rock it held in the begining.

Explanation:

4 0

3 years ago

the determination of a chemical formula n this experiment, you will use the law of definite proportions to find the chemical for

victus00 [196]

Answer: Strictly a laboratory analysis and can only be done using the data obtained during analysis

Explanation:

To find a solution to this problem, you need to use the data collected during the lab work. A guide could be finding the possible forms of hydrated copper chlorides in reference books. Since it's also a lab work, you can definitely compare your data with lab mates.

The formula CuxCly.zH₂O and its name chloride hydrate already gives you an idea of the possibilities of the value of the integers, hence you can take a good guess for the identity of the unknown salt and calculate the theoretical formular weight for it. From the that you can proceed to also find the mass of water and copper from your lab analysis.

7 0

3 years ago

E. The element 231/90Th decays to 231/91 Pa. Use the laws of conservation of charge and nucleon number to determine the decay pa

svetoff [14.1K]

Answer:

A negatron emission

Explanation:

We know that radioactivity orginates from instability of the nucleus. When the nucleus is unstable, radioactive emissions are produced in the form of any of these rays:

> Alpha particle emisson

>Beta particles

> Gamma rays

These emissions create a balance for a radioactive decay.

In balancing nuclear reactions we make sure that the charges on both sides must be conserved and that the mass number and atomic numbers conserved too. This means that the sum of mass number and atomic numbers on both side of the reaction must be equal.

The nucleons are the protons and neutrons, they add up to give the mass number. The atomic number is the proton number.

For the given radioactive reaction:

²³¹₉₀Th → ²³¹₉₁Pa + ?

From this equation, we see that the mass number is conserved but the atomic number is not.

The mass number is the superscript whereas the atomic number is the subscript.

Let's say the decay produces an emission of a particle denoted by X

²³¹₉₀Th → ²³¹₉₁Pa + ᵃₙX

What would the nature of X be?

For the charges and masses to be conserved, X must have mass number of 0 and an atomic number of -1.

Checking:

Mass number:

231 = 231 + a ( a is the mass number)

a = 231 - 231 = 0

Atomic number:

90 = 91 + n

n = 90- 91 = - 1

With X having a mass number of 0 and an atomic number of -1, we have a beta particle emission. Specifically, a negatron has been emitted.

A negatron is denoted as ⁰₋₁β which perfectly makes the equation conserved and suits the description of X.

The complete equation is thus written as:

²³¹₉₀Th → ²³¹₉₁Pa + ⁰₋₁β + energy

6 0

3 years ago

Which isomer of 1-bromo-2-tert-butylcyclohexane reacts faster when refluxed with potassium tert-butoxide?Draw the structure of t

ivolga24 [154]

Answer:

See explanation and image attached

Explanation:

The reaction of 1-bromo-2-tert-butylcyclohexane with potassium tert-butoxide is an elimination reaction that occurs by E2 mechanism.

The E2 reaction proceeds faster when the hydrogens are in an antiperiplanar position at an angle of 180 degrees.

This is only attainable in the trans isomer of 1-bromo-2-tert-butylcyclohexane. Hence trans 1-bromo-2-tert-butylcyclohexane reacts faster with potassium tert-butoxide

8 0

3 years ago