Question

A $1.85$ mole sample of the ionic compound $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ contains how many moles of the following? A) $Al$ atoms, B) $S$ atoms and C) $A{{l}^{3+}}$

Answer 1

Hint : To determine the answer we should know that moles of atoms combine in a fix ratio to give a mole of molecule. We have the molecular formula of aluminium sulphate. From the molecular formula we can determine the moles of each constituting atom present in one mole of the molecule. So, as we have a mole of an atom in one mole of molecules we can determine the moles of atoms in $x$ mole of molecules.

Complete Step By Step Answer:
The given formula of aluminium sulphate is $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ . From this formula we can say one mole of aluminium sulphate is made up of two moles of aluminium atoms, three moles of sulphur atoms and twelve moles of oxygen atoms. So, if one moles of aluminium sulphate contains two moles of aluminium atoms then seven moles of aluminium sulphate will contain,
The moles of $Al$ atoms: There are two atoms of $Al$ per unit of $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ , so there are two moles of $Al$ per mole of $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ :
$1.85mol\left[ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} \right]\left( \dfrac{2mol(Al)}{1mol\left[ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} \right]} \right)=3.70mol(Al).$
The moles of $S$ atoms:
There are three atoms of S per units of $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ , so there are three moles of $S$ per mole of $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ :
$1.85mol\left[ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} \right]\left( \dfrac{3mol(S)}{1mol\left[ A{{l}_{2}}{{(S{{O}_{4}})}_{3}} \right]} \right)=5.55mol(S).$
The moles of $A{{l}^{3+~}}$ ions: The number of ions of $A{{l}^{3+~}}$ will be the same as the number of atoms of $Al$ , so the answer will be the same as that for a).
Therefore,
(A) $3.70mol(Al).$
(B) $5.55mol(S).$
(C) $3.70mol(Al).$

Note :
The subscript present after an atom shows the number of that atom. The subscript present after a bracket presents the number of all the atoms present inside the bracket. In $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ , $3$ outside the bracket shows the three atoms of sulphur and three atoms of oxygen.