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Karolina [17]
3 years ago
13

Titanium dioxide (TiO2) is used extensively as a white pigment. It is produced from an ore that contains ilmenite (FeTiO3) and f

erric oxide (Fe2O3). The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate [(TiO)SO4 ] and ferrous sulfate (FeSO4 ). Water is added to hydrolyze the titanyl sulfate to H2TiO3, which precipitates, and H2SO4. The precipitate is then roasted, driving off water and leaving a residue of pure titanium dioxide. (Several steps to remove iron from the intermediate solutions as iron sulfate have been omitted from this description.) Suppose an ore containing 24.3% Ti by mass is digested with an 80% H2SO4 solution, supplied in 50% excess of the amount needed to convert all the ilmenite to titanyl sulfate and all the ferric oxide to ferric sulfate [Fe2 (SO4 )3 ]. Further suppose that 89% of the ilmenite actually decomposes. Calculate the masses (kg) of ore and 80% sulfuric acid solution that must be fed to produce 1000 kg of pure TiO2.
Chemistry
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer:

2928kg of ore are required.

2585kg of the 80% H₂SO₄ solution are required.

Explanation:

To solve this question we need first to find the moles of titanium in 1000kg of TiO₂. Keeping in mind the 89% of descomposition we can find the mass of the ore and the mass of the 80% sulfuric acid required:

<em>Moles TiO₂ -Molar mass: 79.866g/mol-:</em>

1x10⁶g * (1mol / 79.866g) = 12521 moles Titanium

In mass -Molar mass Ti: 47.867g/mol-:

12521 moles Titanium * (47.867g / mol) = 599341.4g of Ti.

As the ore contains 24.3% of Ti:

599341.4g of Ti = 599.34kg Ti * (100 / 24.3) = 2606kg ore

As the descomposition is just of 89%:

2606kg ore * (100 / 89) =

<h3>2928kg of ore are required</h3><h3 />

<em>Mass 80% sulfuric acid:</em>

12521 moles Titanium = 12521 moles H₂SO₄ * (100/89) = 14068.5 moles of H₂SO₄ are required.

In an excess of 50% =

14068.5 moles of H₂SO₄ are required * 1.5 = 21102.8 moles of H₂SO₄.

The mass is:

21102.8 moles of H₂SO₄ * (98g / mol) = 2068075g = 2068kg of sulfuric acid

That is in the 80%:

2068kg of sulfuric acid * (100/ 80) =

<h3>2585kg of the 80% H₂SO₄ solution are required</h3>
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The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of
Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

C

.

Water has a specific heat of approximately  

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

C

, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

times as much heat to increase its temperature by  

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

5 0
2 years ago
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