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Heat of Reaction

According to ...

Question

According to the equation, the energy evolved when $7.8 g$ benzene is burnt in air will be:
$C_{6} H_{6} + \frac{15}{2} O_{2} (g) \to 6 {C O}_{2} (g) + 3 H_{2} O (l)$ ; $Δ H = - 3264.4 k J$ ${m o l}^{- 1}$

163.22 k J

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32.64 k J

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3.264 k J

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326.4 k J

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Solution

The correct option is C $326.4 k J$
Mass of benzene $= 7.8 g m$
So, $moles of benzene =$ $\frac{mass of benzene}{molecular weight of benzene}$
$m o l e s = \frac{7.8}{78} = 0.1$
Now, as per the given reaction the $Δ H = - 3264.4 K J$ $m o l^{- 1}$ means when 1 mole of benzene reacts than energy evolved $= 3264.4 K J$ .
So, when 0.1 moles react than energy evolved $= 0.1 \times 3264.4 = 326.44 K J$
Hence, answer is option $D$ .

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Similar questions

C_{6} H_{6 (l)} + \frac{15}{2} O_{2 (g)} \to 3 H_{2} O_{(i)} + 6 C O_{2 (g)}

Δ H = - 3264.4 k J / m o l

C_{6} H_{6 (l)} + \frac{15}{2} O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}, Δ H = - 3264

3.9

C_{6} H_{6} (l) + 7 \frac{1}{2} O_{2} (g) \to 3 H_{2} O (l) + 6 C O_{2} (g), Δ H = - 781.0 K c a l m o l^{- 1}

156 g o f C_{6} H_{6}

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to {3 H}_{2} O (l) + {6 C O}_{2} (g)

C_{6} H_{6} (l)

H_{2} O (l)

{C O}_{2} (g)

C_{6} H_{6} (l) + 15 / 2 O_{2} (g) \to 6 {C O}_{2} (g) + 3 H_{2} O (l)

Δ U

25^{o} C = - 3268.12 k J

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