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dmitriy555 [2]
3 years ago
7

What is the name of the compound Al2(SO3)3?

Chemistry
2 answers:
jek_recluse [69]3 years ago
6 0
Take note that this is S03 and not S04.

S03:Sulfite
S04:Sulfate

So the answer  would be Aluminum Sulfite =)


faust18 [17]3 years ago
3 0

Answer:

Name: Aluminum Sulfite

Explanation:

Systematic nomenclature: dialuminium tris [trioxosulfate (IV)]

Oxoanions: anions derived from removing H + from an oxo acid with a vulgar name:

(SO3)3^-2 (sulfite)

Metal cation according to valence number:

Al (III) or Al+3 (as a cation) (Aluminum)

There are different types of salts, in this case we have a simple salt and the nomenclature for binary substances is used when only one class of cation and one class of anion are present

So:

Al2(SO3)3: Aluminum Sulfite

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Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti
djverab [1.8K]
<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

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