A) \[13.7kcal\]
B) \[27.4kcal\]
C) \[6.85\text{ }kcal\]
D) \[3.425kcal\]
Correct Answer: B
Solution :
1 mole of \[{{H}_{2}}S{{O}_{4}}\]will be completely neutralized by two moles of \[O{{H}^{-}}\] ion producing two moles of \[{{H}_{2}}O\]. Hence, the amount of heat liberated will be \[2\times 13.7=27.4kcal\] \[\underset{1\,mole}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{2\,moles}{\mathop{2O{{H}^{-}}}}\,\xrightarrow{{}}S{{O}_{4}}^{2-}+\underset{2\,moles}{\mathop{2{{H}_{2}}O}}\,+27.4kcal\]
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