Question

The mass (in g) of Kaolin , $(A{l_2}S{i_2}{O_7}.2{H_2}O)$ which is required to prepare $10g$ of $A{l_2}{(S{O_4})_3}.18{H_2}O$ is (as nearest integer) :

Answer 1

Hint : We can solve this problem with the help of the concept of Principle of Atom conservation. According to the Principle of Atom conservation, “The total number of atoms of reactant must equal the number of products’’. It comes from the law of conservation of mass, Thus we can say that; Mass of atoms of elements in reactant = mass of atoms of elements in product or Moles of atoms of elements in reactant = Moles of atoms of elements in product.

Complete answer:

So we can calculate the mass (in g) of Kaolin , $(A{l_2}S{i_2}{O_7}.2{H_2}O)$ which is required to prepare $10g$ of $A{l_2}{(S{O_4})_3}.18{H_2}O$ with the help of this principle.
First of all let's calculate the molar mass of Kaolin and $A{l_2}{(S{O_4})_3}.18{H_2}O$.
Molar mass of Kaolin$(A{l_2}S{i_2}{O_7}.2{H_2}O)$ $= 2 \times 27 + 2 \times 28 + 7 \times 16 + 2 \times 18 = 258$ gram per moles.
Molar mass of Alum$(A{l_2}{(S{O_4})_3}.18{H_2}O)$ $= 2 \times 27 + 3 \times 32 + 12 \times 16 + 18 \times 18 = 666.4$ gram per mole. It is given in the problem that $10g$ of $A{l_2}{(S{O_4})_3}.18{H_2}O$ has to be prepared. So according to the Principle of Atom conservation we know that Moles of atoms of elements in reactant = Moles of atoms of elements in product. Let x grams of Kaolin react with $10$ gram of $A{l_2}{(S{O_4})_3}.18{H_2}O$.
$\begin{gathered} \Rightarrow \dfrac{x}{{258}} = \dfrac{{10}}{{666.4}} \\\ \Rightarrow x = 3.87g \\\ \end{gathered}$
But it is asked that the answer should be in the nearest integer so the nearest integer value of $3.87$ is $4$. Hence answer of this problem is $4$ that means the mass (in g) of Kaolin , $(A{l_2}S{i_2}{O_7}.2{H_2}O)$ which is required to prepare $10g$ of $A{l_2}{(S{O_4})_3}.18{H_2}O$ is $4$ gram.

Note : We have approached this problem with the help of the Principle of Atom conservation. To get the weight of Kaolin we have calculated the number of moles of the reactant and product. On equating them we get the answer which we have converted in integer.