Question
1 gram equivalent of $${H_2}S{O_4}$$ is treated with $$112\,g$$ of $$KOH$$ for complete neutralization. Which of the following statementsis correct?
A.
$$13.7\,kcal$$ of heat is evolved with the formation of $$87\,g$$ of $${K_2}S{O_4},$$ leaving no $$KOH.$$
B.
$$27.4\,kcal$$ of heat is evolved with the formation of $$87\,g$$ of $${K_2}S{O_4},$$ leaving 4 gram equivalent of $$KOH.$$
C.
$$15.7\,kcal$$ of heat is evolved with the formation of 1 gram equivalent of $${K_2}S{O_4},$$ leaving $$56\,g$$ of $$KOH.$$
D.
$$13.7\,kcal$$ of heat is evolved with the formation of $$87\,g$$ of $${K_2}S{O_4},$$ leaving 1 gram equivalent of $$KOH.$$
Answer :
$$13.7\,kcal$$ of heat is evolved with the formation of $$87\,g$$ of $${K_2}S{O_4},$$ leaving 1 gram equivalent of $$KOH.$$
Solution :
\[\underset{\begin{smallmatrix}
98 \\
49
\end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{\begin{smallmatrix}
112 \\
56
\end{smallmatrix}}{\mathop{2KOH}}\,\to \underset{\begin{smallmatrix}
174 \\
87
\end{smallmatrix}}{\mathop{{{K}_{2}}S{{O}_{4}}}}\,+\underset{\begin{smallmatrix}
2\,mole \\
1\,mole
\end{smallmatrix}}{\mathop{2{{H}_{2}}O}}\,\]
$$13.7\,kcal$$ is the heat evolved when $$1\,gev$$ of strong acid is neutralised by $$1\,gev$$ of strong base.