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Andre45 [30]
3 years ago
7

What is the molar mass of Al2(SO4)3​

Chemistry
1 answer:
umka21 [38]3 years ago
4 0

Answer:

The molar mass of Al_2(SO_4)_3​ is 342.145 g/mol

Explanation:

The molar mass  is stated as the “mass per unit amount of substance”of a given chemical compound  

Al_2(SO_4)_3​  is Aluminium sulphate. They easily dissolve in water. It is primarily  used as a 'coagulating agent' in the drinking water purification and also in waste and sewage water treatment plants,  

We know,

Atomic weight of Aluminium =  26.981

Atomic weight of Sulphur= 32.065

Atomic weight of  Oxygen =  15.9994

Now molar mass of  Al_2(SO_4)_3​ (Aluminium sulphate) is

=>2(26.981)+3(32.065+4(15.999))

=>53.962 +3(32.065+63.996)

=>53.962+3(96.061)

=>53.962+288.183

=>342.145

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Answer:

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Explanation:

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2 years ago
The mass of an atom is determined by the number of protons the element has. The more protons, the heavier the element.
alekssr [168]

Answer:

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Explanation:

3 0
3 years ago
Write Balance chemical reaction for preparation of chlorine with or without application heat<br>​
suter [353]

Answer:

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

Explanation:

Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.

It is a very reactive gas and is never found in free state in nature.

Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

The set up does not need to be heated.

3 0
2 years ago
What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

3 0
3 years ago
A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp
frozen [14]

Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y  = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y

Replacing z = 0.25; we have:

0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y

replacing z = 0.25

0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y

0.75 = x + 2y

6 0
2 years ago
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