Anyone really good with Organic chemistry

) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79%

LiRa [457]

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

$\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O$

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

$\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}$

3. Add 35% excess

$\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}$

4. Calculate the theoretical volume of nitrogen

$\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}$

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

4HCl + O₂ + N₂ → 2Cl₂ + 2H₂O

I/L: 4 1.35 5.08 0 0

C/L: -0.85(4) -0.85(1) 0 +0.85(2) +0.85(2)

E/L: 0.60 0.50 5.08 1.70 1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

$\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\$

8 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane is

DanielleElmas [232]

Answer:

6g

Explanation:

Step 1:

The balanced equation for the reaction between gaseous ethane and gaseous oxygen. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Step 2:

Determination of the masses of C2H6 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol

Mass of C2H6 from the balanced equation = 2 x 30 = 60g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 7 x 32 = 224g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 4 x 44 = 176g

From the balanced equation above,

60g of C2H6 reacted with 224g of O2 to produce 176g of CO2.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting reactant as it will be needed to obtain the maximum mass of CO2.

From the balanced equation above,

60g of C2H6 reacted with 224g of O2.

Therefore, 2.71g of C2H6 will react with = (2.71 x 224)/60 = 10.12g of O2.

From the above calculation, we can see that a higher mass of O2 is needed to react with 2.71g of C2H6, therefore, O2 is the limiting reactant.

Step 4:

Determination of the maximum mass of CO2 produced when 2.71 g of ethane is mixed with 7.6 g of oxygen.

The limiting reactant is used to determine the maximum mass.

From the balanced equation above,

224g of O2 produce 176g of CO2.

Therefore, 7.6g of O2 will produce = (7.6 x 176)/224 = 5.97g ≈ 6g of CO2

From the calculations made above, the maximum mass of CO2 produced is 5.97 ≈ 6g

5 0

4 years ago

Addition reactions of alkenes are characterized by _________. A) formation of a bond B) addition of two groups across a double b

mr_godi [17]

Answer:

B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND

Explanation:

Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.

During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.

7 0

3 years ago

Add electron dots and charges for the reaction of Ca and O

Nana76 [90]

The two atoms shown in the equation are CALCIUM and oxygen.

You start off with a neutral calcium atom with a shell of two electrons, a shell of 8 around that, a shell of 8 around that, and a shell containing 2...with no charge. 
20 protons + 20 electrons. 

You also have an oxygen atom with a shell of 2, and a shell of 6...with no charge. 
8 protons + 8 electrons. 

Each ionizes to form a calcium ion with 2 electrons removed (from the outer shell), leaving a +2 charge (20 protons, 18 electrons)... 
and an oxygen ion with 2 electrons added (to the outer shell), leaving a -2 charge (8 protons, 10 electrons). 

Their electrostatic attraction causes them to come together to form an ionic compound of CaO in a crystal lattice.

5 0

4 years ago

Explain the results for the tube in which 1.0 m naoh was added to benzoic acid. write an equation for this, using complete struc

nikklg [1K]

When sodium hydroxide (NaOH) is added to benzoic acid (C $C_{6}H_{5}COOH$ ), sodium benzoate and water ( $H_{2}O$ ) are formed because benzoic acid is weak acid and NaOH is strong base. The reaction is shown in diagram 1. To the tube of sodium benzoate, if 6 m HCl is added, sodium benzoate salt will react with HCl and produce benzoic acid and sodium chloride, NaCl. The reaction is shown in diagram 2.

5 0

3 years ago