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Question: How much mass of aluminium can be obtained from \( 0.1 \) mol of \( \text{A}{{\text{l}}_{\text{2}}}{...

How much mass of aluminium can be obtained from 0.10.1 mol of Al2(SO4)3\text{A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}} ? [Atomic mass of aluminium = 27]
(A) 2.72.7 g
(B) 5.45.4 g
(C) 3.23.2 g
(D) 7.87.8 g

Explanation

Solution

One mole of a substance is defined as the amount of mass of a substance that is equal to the mass of 6.023×10236.023\times {{10}^{23}} particles, where the word “particles” refer to ions, molecules, atoms, and subatomic particles. We shall calculate the mass of the compound and use stoichiometry to find the mass of aluminium.

Complete Step by Step Answer
To solve this problem, we need to find out the molecular weight of Al2(SO4)3\text{A}{{\text{l}}_{\text{2}}}{{\left( \text{S}{{\text{O}}_{\text{4}}} \right)}_{\text{3}}} . The molecular weight of any substance is equal to the sum of the atomic weights of the component elements. Therefore it is equal to [2×27+2(32+(16×4))]=342\left[ 2\times 27+2\left( 32+\left( 16\times 4 \right) \right) \right]=342 g/mol\text{g/mol}
This is the weight of 1 mole of aluminium as one mole of aluminium = the molecular weight of the substance
So 1 mole = 342 g/mol\text{g/mol} ,
0.10.1 mole = 34.234.2 g/mol\text{g/mol}
In 342 g/mol\text{g/mol} of aluminium sulphate, 27 g is the weight of aluminium, therefore
In 34.234.2 of aluminium sulphate, 27342×34.2\dfrac{27}{342}\times 34.2 = 2.72.7 g of aluminium.
Therefore the correct answer is option A.

Note
The number 6.023×10236.023\times {{10}^{23}} , which is also known as the Avogadro’s number, was chosen so that the mass of one mole of a chemical compound is numerically equal to the average mass of one molecule of a compound in Daltons. Thus, for example, one mole of water contains 6.023×10236.023\times {{10}^{23}} molecules of water whose total mass is equal to 18.01518.015 g/mol\text{g/mol} and the mean mass of one molecule of water is equal to 18.01518.015 Dalton.
The term “gram-molecule” was used for one mol of molecules and “gram-atom” for one mole of atoms.
The mole is a convenient way to express the amounts of the reactants and the products that are involved in a chemical reaction.