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snow_tiger [21]
2 years ago
9

Titanium dioxide, TiO₂, reacts with carbon and chlorine to give gaseous TiCl₄: TiO₂+2C+2CI₂−TiCI₄+2CO The reaction of 7.39 kg ti

tanium dioxide with excess C and Cl₂ gives 14.24 kg titanium tetrachloride. Calculate the theoretical yield of TiCl₄ (assuming complete reaction) and its percentage yield.
Chemistry
1 answer:
Dima020 [189]2 years ago
3 0

Answer:

17.57kg of TiCl_{4} and its percentage yield is 81.0%

Explanation:

Through the reaction you can get the theoretical amount of  TiCl_{4} that must be produced.

7.39kgTiO_{2}x\frac{1kmolTiO_{2} }{79.867kgTiO_{2}}x \frac{1kmolTiCl_{4}}{1kmolTiO_{2}}x\frac{189.867kgTiCl_{4} }{1kmolTiCl_{4}}=17.57kgTiCl_{4}

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage

yield=\frac{actual amount}{theoretical amount}x100= \frac{14.24kg}{17.57kg}x100=81.0%

You might be interested in
If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct
MA_775_DIABLO [31]

Answer:

1. sp  = XY₂

2. sp²  = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d  = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their  hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5  

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

5 0
2 years ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
If you find that the hydrogen alpha line in a star’s spectrum occurs at a wavelength of 656.45 nm, what is the star’s approximat
Paladinen [302]

Answer:

The answer to the question is

The star’s approximate radial velocity is 68.52 km/s

Explanation:

To solve the

The formula is

\frac{\lambda-\lambda_{rw}}{\lambda_{rw}} = \frac{v_r}{c} where

v_r = velocity of the star

λ = Star's spectrum wavelength = 656.45 nm

\lambda_{rw} = Rest wavelength = 656.30 nm

c = Speed of light = 299 792 458 m / s

Therefore we have

\frac{656.45-656.30}{656.30} =\frac{v_r}{299 792 458} or v_r = 68518.7699 m/s or 68.52 km/s

8 0
2 years ago
The reaction of benzene with (CH3)3CCH2Cl in the presence of anhydrous aluminum chloride produces principally which of these?
Sedaia [141]

Answer:

See explanation and image attached

Explanation:

When the carbocation is formed by the action of AlCl3 on the (CH3)3CCH2Cl, a primary carbocation is formed. The formation of the carbonation is followed by a 1,2-alkyl shift to give a tertiary carbocation which subsequently adds to the benzene ring as shown in the image attached to this answer.

6 0
2 years ago
A student states, "When the alcohol sample was at a temperature of 500 K, all the particles were moving faster than any of the p
lord [1]

Answer: i rlly dont know srry

Explanation:

7 0
2 years ago
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