Question

What mass of isobutylene is obtained from $37g$ of tertiary butyl alcohol by heating with $20\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ at $363K$, if the yield is $65\%$?
A.$16g$
B.$18.2g$
C.$20g$
D.$22g$

Answer 1

To solve this question, you must recall basic stoichiometric fundamentals. If we know the amount of reactants in a given reaction, then we can determine the amount of products formed in the reaction.

Complete step-by-step answer: Tertiary alcohols undergo dehydration readily when heated with sulphuric acid resulting into the formation of an alkene. It is a basic elimination reaction and more substituted alkene is preferably formed by the Saytzeff’s rule. The reaction proceeds with the formation of a carbocation intermediate and thus occurs readily for tertiary alcohols.
The reaction of tertiary butyl alcohol with sulphuric acid to give is given by:

From the reaction we can see that one mole of isobutylene is prepared by 1 mole of tertiary butanol.
The molecular mass of isobutylene is $56g$ and that of tertiary butyl alcohol is $74g$.
Thus we can say that $74g$ of tertiary butyl alcohol gives $56g$ of isobutylene.
So, $37g$ of tertiary butyl alcohol will give $28g$ of isobutylene.
It is given in the question that the yield of reaction is $65\%$.
So the mass of isobutylene obtained will be $m = 28 \times \dfrac{{65}}{{100}}$
$\therefore m = 18.2g$

The correct answer is B.

Note: In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.