Question

According to equation, $ {{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)\xrightarrow{{}}3{{H}_{2}}O(l)+6C{{O}_{2}}(g) $ $ \Delta \text{H}=-3264.4\text{ }kJ/mol $ . The energy evolved when 7.8 g of benzene is burnt in air will be:

• A. $ 3.264\text{ }kJ/mol $
• B. $ 32.64\text{ }kJ/mol $
• C. $ 326.4\text{ }kJ/mol $
• D. $ 163.22\text{ }kJ/mol $

Answer 1

Answer: (C)

$ \underset{(6\times 12+6\times 1=78)}{\mathop{{{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)}}\,\to 3{{H}_{2}}O(l)+6C{{O}_{2}}(g) $ $ \because $ The energy evolved when 78 g. benzene is burnt = 3264.4 kJ/mol $ \therefore $ The energy evolved when 7.8g benzene is burnt $ =\frac{3264.4\times 7.8}{78}=326.44kJ/mol $