$C_{6} H_{5} O H (g) + O_{2} (g) ⟶ {C O}_{2} (g) + H_{2} O (l)$
Magnitude of volume change if $30$ ml of $C_{6} H_{5} O H (g)$ is burnt with excess amount of oxygen, is:

30

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60

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20

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10

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Solution

The correct option is A $30$ ml
The given equation is

$C_{6} H_{5} O H_{(g)} + O_{2 (g)} \to C O_{2 (g)} + H_{2} O_{(l)}$

After balanced equation is

$C_{6} H_{5} O H_{(g)} + 7 O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}$

Volume of reactant is;
Number of moles of $C_{6} H_{5} O H = 1$

Volume of $C_{6} H_{5} O H = 30 m L$

$∴$ Moles of $O x y g e n = 7$

Volume $= 7 \times 30 = 210 m L$

Volume of products is ;
Moles of $C O_{2} = 6$

Volume $= 6 \times 30 = 180 m L$

Moles of $H_{2} O = 3$

Volume $= 3 \times 30 = 90 m L$
therefore,

$∴$ Change in volume $= Pr o d u c t v o l u m e - r e a c tan t v o l u m e$

$\Rightarrow (180 + 90) - (210 + 30)$

$\Rightarrow 270 - 240$

$\Rightarrow 30 m L$

Hence;
This is the required solution.

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C_{6} H_{5} O H (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)

Magnitude of volume change if

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C_{6} H_{5} O H (g)

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