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Volume of Gases and Number of Moles

Liquid benzen...

Question

Liquid benzene $(C_{6} H_{6})$ burns in oxygen according to $2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of $O_{2}$ at STP are needed to complete the combustion of 39 g of liquid benzene ?

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Solution

$2 C_{6} H_{5} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O$

$156$ gm $\to 480$ gm

So for combustion of $39$ g of $C_{6} H_{5}, O_{2}$ required

$= 480 \times \frac{39}{156}$

$= 120$ gm $O_{2}$

So volume of $= 120$ gm $O_{2}$ at STP $= 22.4 \times \frac{120}{32}$

$= 84$ ltr of oxygen

Therefore $= 84$ ltr of oxygen is what is required to completely burn $39$ g of liquid benzene.

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Similar questions

(C_{6} H_{6})

2 C_{6} H_{6} (l) + 15 O_{2} (g) ⟶ 12 C O_{2} (g) + 6 H_{2} O (g)

O_{2}

39

(C_{6} H_{6})

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)

39 g

(C_{6} H_{6})

2 C_{6} H_{6} (l) + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)

39 g

2 C_{6} H_{6 (l)} + 15 O_{2 (g)} \to 12 C O_{2 (g)} + 6 H_{2} O_{(l)}

39 g

C = 12, O = 16

2 C_{6} H_{6} + 15 O_{2} \to 12 C O_{2} + 6 H_{2} O

39

C_{6} H_{6}

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