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According to equation, C(6)H(6)(l)+15//2...

According to equation, `C_(6)H_(6)(l)+15//2 O_(2)(g)rarr 6CO_(2)(g)+3H_(2)O(l), Delta H= -3264.4 "KJ mol"^(-1)` the energy evolved when 7.8 g benzene is burnt in air will be -

A

163.22 KJ

B

32.64 KJ

C

3.264 KJ

D

326.4 KJ

Text Solution

Verified by Experts

The correct Answer is:
D
`C_(6)H_(6(l))+(15)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l)), Delta H=-3264.4 kJ`
1 mole `C_(6)H_(6) = 78g C_(6)H_(6)` undergoes combustion then `Delta H =-3264.4 kJ`
`therefore` when `7.8g C_(6)H_(6)` undergoes combustion
`Delta H=(3264.4)/(78)xx7.8=326.44 kJ`
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