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2 years ago

What is the theoretical yield of bromobenzene in this reaction when 32.0 g of benzene reacts with 69.3 g of bromine?

Chemistry

1 answer:

sergiy2304 [10]2 years ago

3 0

First calculate the number of moles of bromine and benzene.

Number of moles = $\frac{given mass in g}{molar mass}$

Mass of benzene = $32.0 g$ (given)

Molar mass of benzene= $78.11 g/mol$

Substitute the above values in formula, we get

Number of moles of benzene= $\frac{32.0 g}{78.11 g/mol}$

= $0.409 mol$

Mass of bromine = 69.3 g

Molar mass of bromine = $2\times 79.9 g/mol$ = $159.8 g/mol$

Substitute the above values in formula, we get

Number of moles of benzene= $\frac{69.3 g}{159.8 g/mol}$

= $0.43 g/mol$

Now, ratio of benzene and bromine comes out be 1:1 respectively and limiting reagent is benzene because present less as compared to bromine.

Thus, number of moles of bromobenzene = $0.409 mol$

Theoretical yield = $number of moles \times molar mass$

= $0.409 mol \times 157.02 g/mol$ (molar mass of bromonenzene = 157.02 g/mol)

= $64.22118 g$

Hence, theoretical yield of bromobenzene is $64.22118 g$

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Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H

allochka39001 [22]

Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

$\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

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

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $HCl$ react to give 1 mole of $H_2O$

So, 0.030 moles of $HCl$ react to give 0.030 moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

$\text{ Mass of }H_2O=(0.030moles)\times (18g/mole)=0.54g$

Therefore, the theoretical yield of water formed from the reaction is 0.54 grams.

6 0

3 years ago

Discuss the function of cobalt in action of vitamin B12 and its derivatives and how cobaloximes serve as models for the biologic

S_A_V [24]

Answer:

Cobalt and its derivatives act as a catalyst for some enzymatic functions, in this case, it acts as a cofactor.

Cobaloximes are effective electrocatalysts for the production of hydrogen and, therefore, functional models for hydrogenases.

Explanation:

Cobalt is a metal that participates directly in the synthesis of vitamin B12, necessary for the metabolism of various proteins, with megaloblastic anemia being one of the greatest evidences of a lack.

Its main function is that it integrates the vitamin B12 molecule, also called cyanocobalamin, so that lack of cobalt causes the same symptoms as vitamin B12 deficiency.

Vitamin B12 can only be synthesized by bacteria and for this a cobalt atom is required, in fact adult ruminants can synthesize enough vitamin B12 in the rumen provided they have an adequate supply of cobalt.

It acts as a catalyst for some enzymatic functions, in this case, it acts as a cofactor. Very important for example in the transfer of methyl groups, in this way cobalt, together with folic acid participate in the synthesis of methionine.

The dimethylglioxime (dmgH) complexes to the cobalt to give a series of complexes of Co (III), called cobaloxime, that having 4 positions coordinated to nitrogen in a coplanar way and forming a ring similar to the corrínico nucleus; If another substituent also has an N-heterocyclic nature, it can correctly simulate the B12 system.

5 0

3 years ago

The water-gas shift reaction describes the reaction of carbon monoxide and water vapor to form carbon dioxide and hydrogen (the

sesenic [268]

Answer:

ΔH∘ = - 41.2 KJ

Explanation:

We want to obtain the change in enthalpy for the reaction

CO(g) + H₂O(g) → CO₂(g) + H₂(g) (Main reaction)

And we're given the heat of formation of the reactants and products in the reaction

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

To achieve this, we use the Born-Haber cycle.

The Born-Haber cycle entails writing the change in enthalpy of a reaction as a sum of change in enthalpies of a number of reactions that sum up to give the reaction whose enthalpy we needed from the start.

The main reaction is a sum of a sort of combination of Reactions A, B and C. We find this combination now.

From the reactions whose change in enthalpies are given,

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

Dividing through by 2

CO(g) → C(s) + (1/2)O₂(g) ΔH∘=+110.5kJ (the enthalpy is divided by 2 too)

This reaction becomes (Reaction B)/2

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

Changing the direction of the reaction

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=483.6kJ (the sign on the change in enthalpy changes)

Then, dividing by 2

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=+483.6kJ

H₂O(g) → H₂(g) + (1/2)O₂(g) ΔH∘=241.8kJ (the change in enthalpy is divided by 2 too)

This reaction becomes (-Reaction C)/2

But, now, our main reaction can be written as a sum of these new Reactions,

Main Reaction = (Reaction A) + [(Reaction B)/2] + [(- Reaction C)/2]

C(s) + O₂(g) + CO(g) + H₂O(g) → CO₂(g) + C(s) + (1/2)O₂(g) + H₂(g) + (1/2)O₂(g)

Which gives the main reaction after eliminating the O2 that appears on both sides.

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Hence,

(ΔH∘ for the main reaction) = (ΔH∘ for reaction A) + [(ΔH∘/2) for reaction B) - [(ΔH∘/2) for reaction C)

ΔH∘ = - 393.5 + (221/2) - (-483.6/2) = - 41.2 KJ

4 0

2 years ago

What are all the oceans combined called?

gavmur [86]

they are called The 7 Seas

4 0

3 years ago

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What is the volume of 1.9 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

masya89 [10]

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.

Answer: 43 L

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3 years ago

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